Tag: Karnataka state syllabus class 10

Pythagoras Theorem Exercise 12.1 – Class 10

Pythagoras Theorem – Exercise 12.1 – Questions:

a. Numerical problems based on Pythagoras theorem.

  1. The sides of a right angled triangle containing the right angle are 5 cm and 12 cm, find its hypotenuse.
  2. Find the length often diagonal of a square of side 12 cm.
  3. The length of the diagonal of a rectangular playground is 125m and the length of one side is 75m. find the length of the other side,
  4. In triangle LAW, LAW = 90, LNA = 90 , LW = 26cm, LN = 6 cm and AN = 8 cm. Calculate the length of WA.
  5. A door of width 6 meter has an arch above it having a height of 2m. Find the radius of the arch.
  6. A peacock on a pillar of 9 feet height on seeing a snake coming towards its hole situated just below the pillar will fly tp catch it. Of both poses the same speed, how far from the pillar they are going to meet?

b. Riders based on Pythagoras theorem.

  1. In triangle MGN, MP⊥GN. If MG = a units, MN = b  units, GP = c units and PN = d units. Prove that (a + b)(a – b) = (c + d)(c – d)
  2. In right angled triangle ABC, ABC = 90, BD ⊥ AC. If AB = c units, BC  = a  units, BD  = p units, CA = b  units. Prove that

1/a^2 + 1/c^2 = 1/p^2

  1. Derive the formula for height and area of an equilateral triangle.

Pythagoras Theorem – Exercise 12.1 – Solution:

a. Numerical problems based on Pythagoras theorem.

  1. The sides of a right angled triangle containing the right angle are 5 cm and 12 cm, find its hypotenuse.

Solution:

Pythagoras Theorem Exercise 12.1In triangle ABC,

AC2 = AB2 + BC2

= 122 + 52

= 144 + 25

= 169

AC = √169 = 13 cm

 

  1. Find the length often diagonal of a square of side 12 cm.

Solution:

Pythagoras Theorem Exercise 12.1By Baudhayana Theorem to squares,

square of the diagonal = 2 x square of length of its sides

AC2 = 2 x AB2

= 2 x 122

AC = √(2×122) = 12√2 cm

 

  1. The length of the diagonal of a rectangular playground is 125m and the length of one side is 75m. find the length of the other side.

Solution:

Pythagoras Theorem Exercise 12.1BD = 125m and AD = 75 m

In triangle ABD, by pythagoras theorem,

BD2 = AD2 + AB2

1252 =  752 + AB2

15625 = 5625 + AB2

15625 – 5625 = AB2

10000 = AB2

AB = 100 m

 

  1. In triangle LAW, LAW = 90, LNA = 90 , LW = 26cm, LN = 6 cm and AN = 8 cm. Calculate the length of WA.

Solution:

Pythagoras Theorem Exercise 12.1Given LW = 26 cm, NA  = 8 cm and LN = 6 cm

In triangle LNA, by Pythagoras theorem,

LA2 = LN2 + NA2

LA2 = 36 + 64

= 100

LA = 10 cm

In triangle LAW, by Pythagoras theorem

262 = 102 + WA2

676 = 100 + WA2

576 = WA2

WA = √576 = 24 cm

 

  1. A door of width 6 meter has an arch above it having a height of 2m. Find the radius of the arch.

Solution:

Pythagoras Theorem Exercise 12.1LetQ be the centre o the arch. join AQ and PQ.

Let PQ = x cm and AQ = QB = x + 2

In triangle QAP, by Pythagoras theorem,

AP2 = AQ2 + PQ2

(x + 2)2 = x2 + 32

x2 + 4x + 4 = x2 + 9

4x + 4 = 9

4x = 9 – 4 = 5

x = 5/4 = 1.25 cm

Therefore, radius of the arch = x + 2 = 1.25 + 2 = 3.25 cm

 

  1. A peacock on a pillar of 9 feet height on seeing a snake coming towards its hole situated just below the pillar will fly to catch it. Of both poses the same speed, how far from the pillar they are going to meet?

Solution:

Pythagoras Theorem Exercise 12.1Given, height of the pillar PQ = 9 feet and distance of QS = 27 feet

Therefore, RQ = 27 – x

In triangle PQR, by Pythagoras theorem,

PR2 = PQ2 + QR2

x2 = 92 + (27 – x)2

x2 = 81 + 729 – 54x + x2

54x = 81 + 729 = 810

x = 810/54 = 15 feet

Therefore QR = 27 – x = 27 – 15 = 12 feet.

 

b.Riders based on Pythagoras theorem.

  1. In triangle MGN, MP⊥GN. If MG = a units, MN = b  units, GP = c units and PN = d units. Prove that (a + b)(a – b) = (c + d)(c – d)

Solution:

Pythagoras Theorem Exercise 12.1Given, in triangle MGN, MP⊥GN. If  MG  = a units, MN = b  units, GP = c units and PN = d units.

In  triangle MPG, by Pythagoras theorem,

MG = MP  + GP

a = MP + c ————(1)

In  triangle MPN, by Pythagoras theorem,

MN = MP +  PN

b = MP + d —————(2)

subtract (1) from (2),

a2 – b2 = (MP + c)2 – (MP + d)2

a2 – b2 =  MP2 – MP2 + c2 – d2

a2 – b2 = c2 – d2

(a + b)(a – b) =  (c + d)(c – d)

 

  1. In right angled triangle ABC, ABC = 90, BD ⊥ AC. If AB = c units, BC  = a  units, BD  = p units, CA = b  units. Prove that

1/a^2 + 1/c^2 = 1/p^2

Solution:

Pythagoras Theorem Exercise 12.1Given, in right angled triangle ABC, ABC = 90, BD ⊥  AC. If AB  = c units, BC  = a  units, BD  = p units, CA = b  units.

Proof:

BD2 = AD x DC

p2 = a x DC

BC2 = AC x DC

a2 = b x DC

AB2 = AC x AD

c2 = b x AD

Now, 1/a^2 + 1/c^2 = 1/p^2

= 1/b [ 1/DC + 1/AD]

= 1/b [ AD + DC/DC.AD]

= 1/bAC/p^2]

= 1/b [b/p^2]

= 1/p^2

 

  1. Derive the formula for height and area of an equilateral triangle.

Solution:

Pythagoras Theorem Exercise 12.1AB=AC=a, BP=PC=a2, AP=h,

AP perpendicular to BC.

PROOF:

In triangle APC, by Pythagoras Theorem,

AC= AP2 + AC2

a = h2 + (a2 ) 2

a2 = h2 + (a4) 2

a 2– (a4 )2 = h2

4a2a^24 = h2

3a^24 = h

h = √(3𝑎^24) = a/2√3

Now area of triangle = 1/2 x b x h

= 1/2 x a x a/2 √3

= √3 a/4

Quadratic Equations Exercise 9.9 – Class X

Quadratic Equations – Exercise 9.9 – Question:

  1. Draw the graphs of the following quadratic equations:

(i) y = – x2

(ii) y = 3x2

(iii) y = 1/2 x2 – 2

(iv) y = x2 – 2x

(v) y = x2 – 8x + 7

(vi) y = (x + 2)(2 – x)

(vii) y = x2 + x – 6

(viii) y = x2 – 2x + 5

Quadratic Equations – Exercise 9.9 – Solution:

  1. Draw the graphs of the following quadratic equations:

(i) y = – x2

Solution:

x-3-2-10123
y-9-4-10-1-4-9

Quadratic Equations Exercise 9.9

 (ii) y = 3x2

Solution:

x-2-1012
y1230312

Quadratic Equations Exercise 9.9

 

 (iii) y = 1/2 x2 – 2

Solution:

x-2-1012
y03/2-23/20

Quadratic Equations Exercise 9.9

 

 (iv) y = x2 – 2x = x(x – 2)

Solution:

x-2-10123
y830-103

Quadratic Equations Exercise 9.9

 (v) y = x2 – 8x + 7

Solution:

x-10123456789
y1670-5-8-9-8-50716

Quadratic Equations Exercise 9.9
(vi) y = (x + 2)(2 – x)

Solution:

y = (x + 2)(2 – x) = 2x – x2 + 4 – 2x = -x2 + 4

Therefore, y = -x2 + 4

x-3-2-10123
y-503430-5

Quadratic Equations Exercise 9.9

 

 (vii) y = x2 + x – 6

Solution:

x-4-3-2-10123
y60-4-6-6-406

Quadratic Equations Exercise 9.9

 (viii) y = x2 – 2x + 5

Solution:

x-2-101234
y138545813

Quadratic Equations Exercise 9.9

Quadratic Equation – Exercise 9.6 – Class X

So far we have learnt to find the roots of given quadratic equations by different methods. We see that the roots are all real numbers.

Is it possible to determine the nature of roots of a given quadratic equation, without actually finding them? Now, let us learn about this.

Nature of the roots of quadratic equations -Quadratic Equation:

Study the following examples:

  1. Consider the quadratic equation x2 – 2x + 1 = 0

This is in the form of ax2 + bx + c = 0; a = 1 , b = -2 and c = 1

Quadratic Equations - Exercise 9.4 - Class x

  1. Consider the equation x2 – 2x – 3 = 0

This is in the form ax2 + bx + c = 0, where a = 1 , b = -2 , c = -3

2

  1. Consider the quadratic equation x2 – 2x + 3 = 0

This is in the form ax2 + bx + c = 0, where a = 1 , b = -2 , c = 3

3

From the above examples, it is evident that the roots of a quadratic equation can be real and equal, real and distinct or imaginary.

Also, observe that the value of b2 – 4ac determines the nature of the roots. We say the nature of roots depends on the values of b2 – 4ac.

The value of the expression b2 – 4ac discriminates the nature of the roots of ax2 + bx + c = 0 and so it is called the discriminant  of the quadratic equation. It is denoted by symbol ∆ and real as delta.

In general, the roots of the quadratic equation ax2 + bx + c = 0 areQuadratic Equation - Exercise 9.6 - Class X

The above results are presented in the table given below:

DiscriminantNature of roots
∆ = 0real and equal
∆ > 0real and distinct
∆ < 0no real roots(imaginary roots)

Example 1: Determine the nature of the roots of the equation 2x2 – 5x – 1 = 0

Solution:

This is in the form ax2 + bx + c = 0. The coefficient are a = 2, b = -5, c = -1

∆ = b2 – 4ac = (-5)2 – 4(2)(-1) = 25 + 8 = 33


Example 2: Determine the nature of the roots of the eqation 4x2 – 4x + 1 = 0

Solution:

Consider the equation 4x2 – 4x + 1 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 4, b = -4, c = 1

∆ = b2 – 4ac = (-4)2 – 4(4)(1) = 16 – 16 = 0

Therefore, the roots of are real and equal.


Example 3: For what positive values of m, roots of the equation x2 + mx + 4 = 0 are (i) equal (ii) distinct

Solution:

Consider the equation x2 + mx + 4 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = m, c = 4

∆ = b2 – 4ac

∆ = m2 – 4(1)(4)

∆ = m2 – 16

(i) If roots are equal then ∆ = 0

m2 – 16 = 0

m2 = 16

m = ±4

(ii) If roots are distinct, then ∆ > 0

m2 – 16 >0

m2 > 16

m > √16

m > 4


Quadratic equation – Exercise 9.6 – Class X

  1. Discuss the nature of roots of the following equations:

(i) y2 – 7y + 2 = 0

(ii) x2 – 2x + 3 = 0

(iii) 2n2 + 5n – 1 = 0

(iv)  a2 + 4a + 4 = 0

(v) x2 + 3x – 4 = 0

(vi) 3d2 – 2d + 1 = 0

2. For what positive values of m roots of the following equation are

a) Equal b) Distinct c) Imaginary

(i) a2 – ma + 1 = 0

(ii) x2  – mx + 9 = 0

(iii) r2 – (m + 1)r + 4 = 0

(iv) mk2 – 3k + 1 = 0

3. Find the value of ‘p’ for which the quadratic equations have equal roots.

(i) x2 – px + 9 = 0

(ii) 2a2 + 3a + p = 0

(iii) pk2 – 12k + 9 = 0

(iv) 2y2 – py + 1 = 0

(v) (p + 1)n2 + 2(p + 3)n + (p + 8) = 0

(vi) (3p + 1)c2 + 2(p + 1)c + p = 0


Quadratic equation – Exercise 9.6 – Class X

  1. Discuss the nature of roots of the following equations:

(i) y2 – 7y + 2 = 0

Solution:

Consider the equation y2 – 7y + 2 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -7, c = 2

∆ = b2 – 4ac

∆ = (-7)2 – 4(1)(2)

∆ = 49 – 8

∆ = 41

Hence, ∆ = 41 , therefore, ∆ > 0

Therefore, the roots of the equation y2 – 7y + 2 = 0 are real and distinct.


 (ii) x2 – 2x + 3 = 0

Solution:

Consider the equation x2 – 2x + 3 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -2, c = 3

∆ = b2 – 4ac

∆ = (-2)2 – 4(1)(3)

∆ = 4 – 12

∆ = -8

Hence, ∆ = -8 , therefore, ∆ < 0

Thus, the roots of the equation x2 – 2x + 3 = 0 are imaginary.


 (iii) 2n2 + 5n – 1 = 0

Solution:

Consider the equation 2n2 + 5n – 1 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 2, b = 5, c = -1

∆ = b2 – 4ac

∆ = (5)2 – 4(2)(-1)

∆ = 25 + 8

∆ = 33

Hence, ∆ = 33 , therefore, ∆ > 0

Therefore, the roots of the equation 2n2 + 5n – 1 = 0 are real and distinct.


 (iv)  a2 + 4a + 4 = 0

Solution:

Consider the equation a2 + 4a + 4 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = 4, c = 4

∆ = b2 – 4ac

∆ = (4)2 – 4(1)(4)

∆ = 16 – 16

∆ = 0

Hence, ∆ = 0

Therefore, the roots of the equation a2 + 4a + 4 = 0 are real and equal


 (v) x2 + 3x – 4 = 0

Solution:

Consider the equation x2 + 3x – 4 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = 3, c = -4

∆ = b2 – 4ac

∆ = (3)2 – 4(1)(-4)

∆ = 9 + 16

∆ = 25

Hence, ∆ = 25 , therefore, ∆ > 0

Therefore, the roots of the equation x2 + 3x – 4 = 0 are real and distinct.


 (vi) 3d2 – 2d + 1 = 0

Solution:

Consider the equation 3d2 – 2d + 1 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 3, b = -2, c = 1

∆ = b2 – 4ac

∆ = (-2)2 – 4(3)(1)

∆ = 4 – 12

∆ = -8

Hence, ∆ = -8 , therefore, ∆ < 0

Therefore, the roots of the equation 3d2 – 2d + 1 = 0 are imaginary.


2. For what positive values of m roots of the following equation are

a) Equal b)Distinct c)Imaginary

(i) a2 – ma + 1 = 0

Solution:

Consider the equation a2 – ma + 1 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -m, c = 1

∆ = b2 – 4ac

∆ = (-m)2 – 4(1)(1)

∆ = m2 – 4

(i) If roots are equal then ∆ = 0

m2 – 4 = 0

m2 = 4

Therefore, at m = 2 the roots of the quadratic equation a2 – ma + 1 = 0 are equal.

(ii) If roots are distinct, then ∆ > 0

m2 – 4 >0

m2 > 4

m > √4

⸫ m > +2

(iii) If roots are imaginary, then ∆ < 0

m2 – 4 < 0

m2 < 4

m < √4

⸫ m < -2

 

 (ii) x2  – mx + 9 = 0

Solution:

Consider the equation x2 – mx + 9 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -m, c = 9

∆ = b2 – 4ac

∆ = (-m)2 – 4(1)(9)

∆ = m2 – 36

(i) If roots are equal then ∆ = 0

m2 – 36 = 0

m2 = 36

m = 6

(ii) If roots are distinct, then ∆ > 0

m2 – 36 >0

m2 > 36

m > √36

⸫ m > +6

(iii) If roots are imaginary, then ∆ < 0

m2 – 36 < 0

m2 < 36

m < √36

⸫ m < -6

 

 (iii) r2 – (m + 1)r + 4 = 0

Solution:

Consider the equation r2 – (m + 1)r + 4 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -(m+1), c = 4

∆ = b2 – 4ac

∆ = [-(m+1)]2 – 4(1)(4)

∆ = (m + 1)2 – 16

∆ = m2 + 2m + 1 – 16

∆ = m2 + 2m – 15

(i) If roots are equal then ∆ = 0

m2 + 2m – 15 =0

m2 + 5m – 3m – 15 = 0

m(m + 5)-3(m + 5) = 0

(m – 3)(m + 5) = 0

m – 3 = 0 or m + 5 = 0

m = 3  or m = -5

Therefore, at no value m can be the quadratic equation mk2 – 3k + 1 = 0 which has equal roots.

(ii) If roots are distinct, then ∆ > 0

m2 + 2m – 15 >0

m2 – 3m + 5m – 15 > 0

m(m – 3)-1(m +3) > 0

(m + 3)(m – 1) > 0

m + 3 > 0 or m – 1 > 0

⸫ m > 1

(iii) If roots are imaginary, then ∆ < 0

m2 + 2m – 3 < 0

m2 +3m – m – 3 < 0

m(m + 3)-1(m +3) < 0

(m + 3)(m – 1) < 0

m + 3 < 0 or m – 1 < 0

⸫ m < -3

 

 (iv) mk2 – 3k + 1 = 0

Solution:

Consider the equation mk2 – 3k + 1 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = m, b = -3, c = 1

∆ = b2 – 4ac

∆ = [-3]2 – 4(m)(1)

∆ = 9 – 4m

(i) If roots are equal then ∆ = 0

9 – 4m =0

-4m = – 9

m = 9/4

Therefore, at m = 9/4 the quadratic equation mk2 – 3k + 1 = 0 has equal roots

(ii) If roots are distinct, then ∆ > 0

9 – 4m > 0

-4m > – 9

⸫ m > 9/4

(iii) If roots are imaginary, then ∆ < 0

9 – 4m < 0

-4m < – 9

⸫ m < 9/4


3.. Find the value of ‘p’ for which the quadratic equations have equal roots.

(i) x2 – px + 9 = 0

Solution:

x2 – px + 9 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -p, c = 9

∆ = (-p)2 – 4x1x9

∆ = [p]2 – 36

∆ = p2 – 36

If roots of quadratic equation x2 – px + 9 = 0 then ∆ = 0

p2 – 36 = 0

p2 = 36

p = √36 = 6


 (ii) 2a2 + 3a + p = 0

Solution:

2a2 + 3a + p = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 2, b = 3, c = p

∆ = (3)2 – 4x2xp

∆ = [3]2 – 8p

∆ = 9 – 8p

If roots of quadratic equation 2a2 + 3a + p = 0 then ∆ = 0

9 – 8p = 0

9 = 8p

p = 9/8


 (iii) pk2 – 12k + 9 = 0

Solution:

pk2 – 12k + 9 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = p, b = -12, c = 9

∆ = b2 – 4ac

∆ = [-12]2 – 4(p)(9)

∆ = 144 – 36p

If roots of quadratic equation pk2 -12k + 9 = 0 then ∆ = 0

144 – 36p = 0

36p = 144

p = 144/36 = 24/6 = 4

Therefore, p = 4 .


 (iv) 2y2 – py + 1 = 0

Solution:

2y2 – py + 1 = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = 2, b = -p, c = 1

∆ = b2 – 4ac

∆ = [-p]2 – 4(2)(1)

∆ = p2 – 8

If roots of quadratic equation 2y2 – py + 1 = 0 then ∆ = 0

p2 – 8 = 0

p2 = 8

p = √8 = 2√2

Therefore, p = 2√2


 (v) (p + 1)n2 + 2(p + 3)n + (p + 8) = 0

Solution:

Consider the quadratic equation (p + 1)n2 + 2(p + 3)n + (p + 8) = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = (p + 1), b = 2(p + 3), c = (p + 8)

∆ = b2 – 4ac

∆ = [2(p+3)]2 – 4(p+1)(p+8)

∆ = 4(p + 3)2 – 4[p2 + 8p + p + 8]

∆ = p2 + 6p + 9 – p2 – 9p – 8

∆ =-3p + 1

If roots of quadratic equation (p + 1)n2 + 2(p + 3)n + (p + 8) = 0 then ∆ = 0

-3p + 1 = 0

-3p = – 1

p = 1/3


 (vi) (3p + 1)c2 + 2(p + 1)c + p = 0

Solution:

Consider the quadratic equation (3p + 1)c2 + 2(p + 1)c + p = 0

This is in the form ax2 + bx + c = 0. The coefficient are a = (3p + 1), b = 2(p + 1), c = p

∆ = b2 – 4ac

∆ = [2(p + 1)]2 – 4(3p+1)p

∆ = 4[p2 + 2p + 1] – 12p2 – 4p

∆ = 4p2 + 8p + 4 – 12p2 – 4p

∆ = -8p2 + 4p + 4

∆ = 8p2 – 4p – 4

If roots of quadratic equation (3p + 1)c2 + 2(p + 1)c + p = 0 then ∆ = 0

8p2 – 4p – 4 = 0

8p2 – 8p + 4p – 4 = 0

8p(p – 1)+4(p – 1) = 0

(8p + 4)(p – 1) = 0

8p + 4 = 0 or p – 1 = 0

8p = -4 or p = 1

p = –4/8 or p = 1

Therefore, p = –1/­2 or p = 1


Quadratic Equations – Exercise 9.1 – Class X

Quadratic Equations – Exercise 9.2 – Class X

Quadratic Equations – Exercise 9.3 – Class X

Quadratic Equations – Exercise 9.4 – Class X

Quadratic Equations – Exercise 9.5 – Class X


 

Quadratic Equations – Exercise 9.5 – Class X

We know that, in mathematics calculations and solving problems are made easier by using formulae. In the same way, quadratic equation can be easily solved by using a formula. The quadratic formula, which is very useful for finding its roots can be derived using the method of completing the square. Let us derive the quadratic formula and learn how to use it for finding roots of the quadratic equations.

(c) Solution of a quadratic equation by formula method:

Consider the quadratic equation ax2 + bx + c = 0, a≠0

Divide the equation by a (i.e., coefficient of x2)

x2 + b/a x + c/a = 0

Find half the coefficient of x and square it,

1/2 x b/a = b/2a

(b/2a)2 = b^2/(4a)^2

Transpose the constant c/a to RHS

2 + b/a x = – c/a

Add (b/2a)to both sides of the equation

2 + b/a x + (b/2a)2 =(b/2a)2c/a

Factorise LHS and simplify RHS

(x + b/2a)2 = b^2/4a^2c/2a

Take square root on both sides of the equation

x + b/2a = ± √(b^2  – 4ac/4a^2) = ±√(b^2 – 4ac)/2a­

x = –b/2a ±√(b^2 – 4ac)/2a­

x = -b±√(b^2 – 4ac)/2a

Therefore, the roots of the quadratic equation ax2 + bx + x = 0 are –b – √(b^2 – 4ac)/2a and –b + √(b^2 – 4ac)/2a

x = -b±√(b^2 – 4ac)/2a is known as quadratic formula.


Example: Solve the quadratic equation x2 – 7x + 12 = 0 by formula method

Solution:

Given x2 – 7x + 12 = 0

The given quadratic equation x2 – 7x + 12 = 0  is of the form ax2 + bx + c = 0 where a = 1, b = – 7 and c = 12

Quadratic Equations - Exercise 9.5


Quadratic Equation – Exercise 9.5 – Class X

Solve the following quadratic equations by using the formula method.

  1. x2 – 4x + 2 = 0
  2. x2 – 2x + 4 = 0
  3. 2y2 + 6y = 3
  4. 15m2 – 11m + 2 = 0
  5. 8r2 = r + 2
  6. (2x + 3)(3x – 2) + 2 = 0
  7. a(x2 + 1) = x(a2 + 1)
  8. x2 + 8x + 6 = 0

Quadratic Equation – Exercise 9.5 – Solution:

Solve the following quadratic equations by using the formula method.

  1. x2 – 4x + 2 = 0

Solution:

x2 – 4x + 2 = 0

Quadratic Equations - Exercise 9.5


  1. x2 – 2x + 4 = 0

Solution:

x2 – 2x + 4 = 0

Quadratic Equations - Exercise 9.5


  1. 2y2 + 6y = 3

Solution:

2y2 + 6y – 3 = 0

Quadratic Equation - Exercise 9.5


  1. 15m2 – 11m + 2 = 0

Solution:

15m2 – 11m + 2 = 0

Quadratic Equations - Exercise 9.5


  1. 8r2 = r + 2

Solution:

8r2 – r – 2 = 0

Quadratic Equations - Exercise 9.5


  1. (2x + 3)(3x – 2) + 2 = 0

Solution:

(2x + 3)(3x – 2) + 2 = 0

6x2 – 4x + 9x – 6 + 2 = 0

6x2 + 5x – 4 = 0

Quadratic Equations - Exercise 9.5


  1. a(x2 + 1) = x(a2 + 1)

Solution:

a(x2 + 1) = x(a2 + 1)

ax2 + a = a2x + x

ax2 – a2x – x + a = 0

ax2 –x(a2 + 1) + a = 0

Quadratic Equations - Exercise 9.5


  1. x2 + 8x + 6 = 0

Solution:

x2 + 8x + 6 = 0

Quadratic Equations - Exercise 9.5


Quadratic Equations – Exercise 9.1 – Class X

Quadratic Equations – Exercise 9.2 – Class X

Quadratic equations – Exercise 9.3 – Class X

Quadratic Equations – Exercise 9.4 – Class X


 

Quadratic Equations – Exercise 9.4 – Class X

Solving Quadratic Equations by Completing the Square Method:

Let us consider the quadratic equation by completing the square equation, x2 + 5x + 5 = 0. Here in the middle tem 5x cannot split into terms such that m + n = 5 and mn = 5. This means we cannot resolve the equation as product of two factors and therefore, it cannot be solved by factorisation method. This is the limitation of factorisation method of solving quadratic equation by factorisation. This method can be used only when it  is possible to split the middle term and factorise  the given quadratic polynomial.

Then, how to solve the quadratic equation where the quadratic polynomial cannot be factorised?

Let us take an example

Example : Solve the quadratic equation 3x2 – 5x + 2 = 0 by completing the square method 

Solution:

Given 3x2 – 5x + 2 = 0.

Here, the coefficient of x2 is 3 and it is not a perfect square.

Such quadratic equations are solved in two ways. Let us do both of them.

(i) Multiply the equation throughout by 3.

(3x2 – 5x + 2 = 0)x3

9x2 – 15x + 6 = 0

Now, half of the coefficient of x is 5/2 .

b = 5/2 and b2 = (5/2)2

So,  9x2 – 15x + (5/2)2 – (5/2)2 + 6 = 0

(3x)2 – 2(3x)( 5/2) + (5/2)2 = (5/2)2 – 6

(3x)2 – 2(3x)(5/2) + (5/2)2 = 25/4 – 6

(3x –  5/2)2 = 25 –  24/4

(3x –  5/2) = ± 1/2

3x  = ± 1/2 + 5/2

3x =  1/2 + 5/2 or 3x = – 1/2 + 5/2

3x =  1+5/2  or 3x = -1+5/2

3x = 6/2 or 3x = 4/2

x = 6/2 x 1/3 = 1 or x = 4/2 x 1/3

x = 1 or x = 4/6 = 2/3

Therefore, 1 and 2/3 are the roots of the quadratic equation 3x2 – 5x + 2 = 0

 (ii) Dividing the equation throughout by 3.

(3x2 – 5x + 2 = 0) ⁄3

Now, let us proceed as earlier

{b = 1/2 x 5/3 = 5/6 ; b2 = (5/6)2}

So, x – 5/3 x + (5/6)2 – (5/6)2 + 2/3 = 0

(x – 5/6)2 = (5/6)22/3

(x – 5/6)2 = 25/362/3

(x – 5/6)2 = 25-24/36

(x – 5/6)2 = 1/36

(x – 5/6) = ± 1/6

x = ± 1/6 + 5/6

x = +1/6 +5/6 or x = –1/6 +5/6

x = 6/6 or x = 4/6

x = 1 or x = 2/3

Therefore, 1 and 2/3 are the roots of the quadratic equation 3x2 – 5x + 2 = 0


Quadratic Equations – Exercise 9.4 – Class 10

1. Solve the following quadratic equations by completing the square.

(i) 4x2 – 20x + 9 = 0

(ii) 4x2 + x – 5 = 0

(iii) 2x2 + 5x – 3 = 0

(iv) x2 + 16x – 9 = 0

(v)x2 – 3x + 1 = 0

(vi) t2 + 3t = 7


Quadratic Equations – Exercise 9.4 – Solutions:

Solve the following quadratic equations by completing the square.

(i) 4x2 – 20x + 9 = 0

Solution:

4x2 – 20x + 9 = 0

4x2 – 18x – 2x + 9 = 0

2x(2x – 9) – 1(2x – 9) = 0

(2x – 1)(2x – 9) = 0

2x – 1 = 0 or 2x – 9 = 0

2x = 1 or 2x = 9

x = 1/2 or x = 9/2

Therefore, 1/2 and 9/2 are the roots of  the quadratic equation 4x2 – 20x + 9 = 0


 (ii) 4x2 + x – 5 = 0

Solution:

4x2 + x  – 5 = 0

4x2 + 5x – 4x  – 5 = 0

x(4x + 5) – 1(4x + 5) = 0

(4x + 5)(x – 1) = 0

4x + 5 = 0 or x – 1 = 0

4x = -5 or x = 1

x = –5/4 or x = 1

Therefore, 1 and –5/4 are the roots of  the quadratic equation 4x2 + x  – 5 = 0


 (iii) 2x2 + 5x – 3 = 0

Solution:

2x2 + 5x – 3 = 0

2x2 + 6x – x – 3 = 0

2x(x + 3) – 1(x + 3) = 0

(2x – 1)(x + 3) = 0

2x – 1 = 0 or x + 3 = 0

2x = 1 or x = -3

x = 1/2 or x = -3

Therefore, x = 1/2 or – 3 is the root of the Quadratic equation  2x2 + 5x – 3 = 0.


 (iv) x2 + 16x – 9 = 0

Solution:

x2 + 16x – 9 = 0

Half of the coefficient of x is 16/2 . Therefore,  b = 8   and b2 = (8)2 = 64

x2 + 16x + (16/2)2 – (16/2)2– 9 = 0

x2 + 16x + (16/2)2 = (16/2)2 + 9

x2 + 2(8)(x) + 82 = (16/2)2 + 9

(x + 8)2 = 82 + 9

(x + 8)2 = 64 + 9

(x + 8) = ±√73

x = – 8 ±√73

Therefore, x = – 8 ±√73 is the root of the Quadratic equation x2 + 16x – 9 = 0


(v)x2 – 3x + 1 = 0

Solution:

x2 – 3x + 1 = 0

Half of the coefficient of x is 3/2 . Therefore,  b = 3/2   and b2 = (3/2)2

x2 – 3x + (3/2)2 – (3/2)2 + 1 = 0

x2 – 2(x)(3/2) + (3/2)2 = (3/2)2 – 1

(x – 3/2)2 = 9/4 – 1

(x – 3/2)2 = 9 – 4/4

(x – 3/2)2 = 5/4

(x – 3/2) = ±√(5/4)

x = ±√(5/4) + 3/2

x = 3/2 ±5/2

x = 3±√5/2

Therefore, x = 3±√5/2 is the root of the quadratic equation x2 – 3x + 1 = 0


(vi) t2 + 3t = 7

Solution:

t2 + 3t = 7

t2 + 3t – 7 = 0

Half of the coefficient of t is 3/2 . Therefore,  b = 3/2  and b2 = (3/2)2

t2 + 3t  + (3/2)2 – (3/2)2 – 7 = 0

t2 + 2(t)( 3/2) + (3/2)2 = (3/2)2 + 7

(t + 3/2)2 = 9/4 + 7

(t + 3/2)2 = 9+28/4

(t + 3/2)2 = 37/4

(t + 3/2) = ±√(37/4 )

(t + 3/2) = ±37/2

t = ±37/23/2

t = ±37 – 3/2

or

t =-3 ± √37/2

Therefore, t = -3 ± √37/2 is the root of the quadratic equation t2 + 3t = 7


Quadratic Equations – Exercise 9.1 – Class X

Quadratic Equations – Exercise 9.2 – Class X

Quadratic equations – Exercise 9.3 – Class X


 

 

 

 

 

 

 

Quadratic equations – Exercise 9.3 – Class X

Solutions of adfected Quadratic equations – Quadratic Equations:

We know the general form of an adfected quadratic equation is ax² + bx + c, a≠0. this equation can also occur in different forms such as ax² + bx = 0, ax² + c = 0, and ax² =  0.

Solution of a quadratic equation by factorisation method – Quadratic Equations:

Factorisation method is used when the quadratic equation can be factorised can be factorised into two linear factors. After factorisation, the quadratic equation is expressed as the product of its two linear factors and this is equated to zero. That is ax² + bx + c = 0 and (x ± m)(x ± n) = 0. Then we apply zero product rule and equate each factor to zero and solve for the unknown.

i.e.,

Quadratic Equations - Exercise 9.3

So,  ±m and  ±n are the roots of the quadratic equation ax² + bx + c = 0.


Quadratic Equations – Exercise 9.3

  1. Solve the quadratic equations by factorization method:

(i) x2 + 15x + 50 = 0

(ii) 6 – p2 = p

(iii) 100x2 – 20x + 1 = 0

(iv) √2x2 + 7x +5√2 = 0

(v)x2 + 4kx + 4k2 = 0

(vi) m – 7/m = 6

(vii) 0.2t2 – 0.04t = 0.03

(viii) √5x2 + 2x= 3√5

(ix) x/x+1 + x+1/x = 34/15

(x) x-1/x-2 + x-3/x-4 = 31/3

(xi) a2b2x2 –(a2 + b2)x + 1 = 0

(xii) (2x – 3) = √(2x2 – 2x + 2)


Quadratic Equations – Exercise 9.3

  1. Solve the quadratic equations by factorisation method:

(i) x2 + 15x + 50 = 0

Solution:

x2 + 15x + 50 = 0

x2 + 10x + 5x + 50 = 0

x(x + 10) + 5(x + 10) = 0

(x + 10)(x + 5) = 0

x = -10 or x = -5


 (ii) 6 – p2 = p

Solution:

6 – p2 – p = 0

p2 + p – 6 = 0

p2 + 3p – 2p – 6 = 0

p(p + 3)-2(p + 3) = 0

(p – 2)(p + 3) = 0

p = 2 or p = -3


 (iii) 100x2 – 20x + 1 = 0

Solution:

(10x)2 – 20x + 1 = 0

100x2 – 10x – 10x + 1 = 0

10x(10x – 1)-1(10x – 1) = 0

(10x – 1)(10x – 1) = 0

10x = 1 or 10x = 1

x = 1/10 or x = 1/10


 (iv) √2x2 + 7x +5√2 = 0

Solution:

√2x2 + 7x +5√2 = 0

√2x2 + 2x + 5x +5√2 = 0

√2x(x + √2) + 5(x + √2) = 0

(√2x + 5)(x + √2) = 0

√2x = -5 or x = -√2

x = -5/√2 or x = -√2


 (v)x2 + 4kx + 4k2 = 0

Solution:

x2 + 4kx + 4k2 = 0

x2 + 2kx + 2kx + 4k2 = 0

x(x + 2k)+2k(x + 2k) = 0

(x + 2k)(2 + 2k) = 0

x = -2k or x =  -2k


 (vi) m – 7/m = 6

Solution:

m – 7/m = 6

m2 – 7 = 6m

m2 – 6m – 7 = 0

m2 – 7m + m – 7 = 0

m(m – 7) +1(m – 7) = 0

(m + 1)(m – 7) = 0

m = -1 or m = 7


 (vii) 0.2t2 – 0.04t = 0.03

Solution:

0.2t2 – 0.04t = 0.03

0.2t2 – 0.04t – 0.03 = 0

20t2 – 4t – 3 = 0

20t2 – 10t + 6t – 3 = 0

10t(2t – 1)+3(2t – 1) = 0

(10t + 3)(2t – 1) = 0

10t + 3 = 0 or 2t – 1 = 0

10t = -3 or 2t = 1

t = -3/10 or t = 1/2


 (viii) √5x2 + 2x= 3√5

Solution:

√5x2 + 2x= 3√5

√5x2 + 2x – 3√5 = 0

√5x2 + 3x – 5x – 3√5 = 0

x(√5x + 3) -√5(√5x + 3) = 0

(x – √5)( √5x + 3) = 0

x – √5 = 0 or √5x + 3 = 0

x = √5 or √5x = – 3

x = √5 or x = -3/√5


 (ix) x/x+1 + x+1/x = 34/15

Solution:

x/x+1 + x+1/x = 34/15

x/x+1 + x+1/x34/15 = 0

x[15x(x+1)]/x+1 + (x+1)[15x(x+1)]/x34[15x(x+1)]/15 = 0

15x2 +  15(x + 1)2 – 34x(x+1) = 0

15x2 + 15(x2 + 2x + 1) – 34x2 –  34x = 0

15x2 + 15x2 +30x + 15 – 34x2 – 34x = 0

-4x2 – 4x +15 = 0

4x2 +  4x – 15 = 0

4x2 – 6x + 10x – 15 = 0

2x(2x – 3) +5(2x – 3)  = 0

(2x + 5)(2x – 3) = 0

2x + 5 = 0 or 2x – 3 = 0

2x = -5 or 2x  = 3

x = –5/2  or x = 3/2


 (x) x-1/x-2 + x-3/x-4 = 31/3

Solution:

x-1/x-2 + x-3/x-4 = 31/3

x-1/x-2 + x-3/x-410/3 = 0

(x-1)(x – 2)(x – 4)3/(x-2) + (x – 3)(x – 2)(x – 4)3/(x – 4)10(x – 2)(x – 4)3/3 = 0

3(x – 1)(x – 4) + 3(x – 3)(x – 2) – 10(x –  2)(x – 4) = 0

3(x2 – 4x – x  + 4) + 3(x2 – 2x – 3x + 6) – 10(x2 – 4x – 2x + 8) = 0

3x2 – 12x – 3x  + 12 + 3x2 – 6x – 9x + 18 – 10x2 + 40x + 20x – 80 = 0

-4x2 + 30x – 50 = 0

4x2 – 30x + 50 = 0

2x2 – 15x + 25 = 0

2x2 – 10x – 5x + 25 = 0

2x(x – 5)-5(x – 5) = 0

(2x – 5)(x – 5) = 0

2x – 5 = 0 or x – 5 = 0

x = 5/2 or x = 5


 (xi) a2b2x2 –(a2 + b2)x + 1 = 0

Solution:

a2b2x2 –(a2 + b2)x + 1 = 0

a2b2x2 –a2x – b2x + 1 = 0

a2x(b2x – 1)-1(b2 x – 1) = 0

(a2x – 1)(b2x – 1) = 0

a2x – 1 = 0 or b2x – 1 = 0

a2x = 1 or b2x = 1

x = 1/a^2 or x = 1/b^2


 (xii) (2x – 3) = √(2x2 – 2x + 21)

Solution:

(2x – 3) = √(2x2 – 2x + 21)

(2x – 3)2 = (2x2 – 2x + 21)

4x2 – 12x + 9 = 2x2 – 2x + 21

4x2 – 12x + 9 – (2x2 – 2x + 21) = 0

4x2 – 12x + 9 – 2x2 + 2x – 21 = 0

2x2 – 10x – 12 = 0

x2 – 5x – 6 = 0

x2 – 6x + x – 6 = 0

x(x – 6) +1(x – 6) = 0

(x – 6)(x + 1) = 0

x – 6 = 0 or x + 1 = 0

x = 6 or x = -1


Quadratic Equations – Exercise 9.1 – Class X

Quadratic equations – Exercise 9.2 – Class X

Quadratic Equations – Exercise 9.2 – Class X

Previous Exercise – Quadratic Equations – Exercise 9.1 – Class X


Quadratic equations – Exercise 9.2

1.Classify the following equations into pure and adfected quadratic equations:

(i) x2 = 100

(ii) x2 + 6 = 100

(iii) p(p – 3) = 1

(iv) x3 + 3 = 2x

(v) (x+ 9)(x – 9) = 0

(vi) 2x2 = 72

(vii) x2 – x = 0

(viii)7x = 35/x

(ix) x + 1/x = 5

(x) 4x = 81/x

(xi) (2x – 5)2 = 81

(xii) (x – 4)^2/18 = 2/9

2.Solve the quadratic equations:

(i)x2 – 196 = 0

(ii) 5x2= 625

(iii) x2 + 1 = 101

(iv) 7x = 64/7x

(v) (x + 8)2 – 5=31

(vi) x^2/23/4 = 71/4

(vii) -4x2 + 324 = 0

(viii) -37.5 x2 = -37.5

3.In each of the following determine whether the given values of variables is a solution of the quadratic equation or not.

(i)x2 + 14x + 13 = 0; x = -1 ; x =-13

(ii) 7x2 – 12x = 0 ; x = 1/3

(iii) 2m2 – 6m + 3 = 0; m = 1/2

(iv) y2 + √2y – 4 = 0 ; y = 2√2

(v) x/x+2 = 1/2 ; x = 2 and x = 1

(vi) 6x2 – x – 2 = 0 ; x = –1/2 and x = 2/3

4. Solve the quadratic equations:

(i)If A= πr2, solve for r and find the value of r if A = 77 and π = 22/7

(ii) If r2 = l2 + d2 solve for d and find the value of d if r = 5 and l = 4

(iii) If A = √3a^2/4 solve for v and find the value of a, if A = 16√3

(iv) if k = 1/2 mv2 solve for v and find the value of v, if k = 100 and m = 2

(vi) If  v2 = u2 + 2as solve for v and find the value of v, if u = 0, a = 2, s = 100


Quadratic equations – Exercise 9.2

1.Classify the following equations into pure and adfected quadratic equations:

(i) x2 = 100

Solution:

The given equation, x2 = 100

x2 – 100 = 0

The given equation is having only second degree variable.

Therefore, the given equation x2 = 100 is pure quadratic equation.


 (ii) x2 + 6 = 100

Solution:

The given equation, x2 + 6 = 100

x2 + 6 – 100 = 0

x2 – 94 = 0

The given equation is having only second degree variable.

Therefore, the given equation x2 + 6 = 100 is pure quadratic equation.


 (iii) p(p – 3) = 1

Solution:

The given equation, p(p – 3) = 1

p2 – 3p – 1 = 0

The given equation is having both second degree and first degree variable.

Therefore, the given equation p(p – 3) = 0 is adfected quadratic equation.


 (iv) x3 + 3 = 2x

Solution:

The given equation, x3 + 3= 2x

x3 – 2x + 3 = 0

The given equation is having third degree variable.

Therefore, the given equation x3 + 3 = 2x is not a quadratic equation.


 (v) (x+ 9)(x – 9) = 0

Solution:

The given equation, (x + 9)(x – 9) = 0

x2 – 9x + 9x – 81 = 0

x2 – 81 = 0

The given equation is having only second degree variable.

Therefore, the given equation (x + 9)(x – 9)= 0 is pure quadratic equation.


 (vi) 2x2 = 72

Solution:

The given equation, 2x2 = 72

2x2 – 72 = 0

The given equation is having only second degree variable.

Therefore, the given equation 2x2 = 72 is pure quadratic equation.


 (vii) x2 – x = 0

Solution:

The given equation, x2 – x = 0

The given equation is having both second degree and first degree variable.

Therefore, the given equation x2 – x = 0 is adfected quadratic equation.


 (viii)7x = 35/x

Solution:

The given equation, 7x = 35/x

7x2 = 35

7x2 – 35 = 0

The given equation is having only second degree variable.

Therefore, the given equation 7x = 35/x is pure quadratic equation.


 (ix) x + 1/x = 5

Solution:

The given equation, x + 1/x = 5

x2 + 1 = 5x

x2 – 5x + 1 = 0

The given equation is having both first degree and second degree variable.

Therefore, the given equation x2 – 5x + 1 = 0 is adfected quadratic equation.


 (x) 4x = 81/x

Solution:

The given equation, 4x = 81/x

4x2 = 81

4x2 – 81 = 0

The given equation is having only second degree variable.

Therefore, the given equation 4x = 81/x is pure quadratic equation.


 (xi) (2x – 5)2 = 81

Solution:

The given equation, (2x – 5)2 = 81

(2x)2 – 2(2x)(5) + 52 = 81

4x2 – 10x + 25 – 81 = 0

4x2 – 10x – 56 = 0

The given equation is having both first degree and second degree variable.

Therefore, the given equation (2x – 5)2 = 81 is adfected quadratic equation.


 (xii) (x – 4)^2/18 = 2/9

Solution:

The given equation, (x – 4)^2/18 = 2/9

9(x – 4)2 = 2 x 18

9(x2 – 8x + 16) = 36

9x2 – 72x + 144 – 36 = 0

9x2 – 72x + 108 = 0

The given equation is having both first degree and second degree variable.

Therefore, the given equation (x – 4)^2/18 = 2/9 is adfected quadratic equation.


2.Solve the quadratic equations:

(i)x2 – 196 = 0

Solution:

x2 – 196 = 0

x2 = 196

x = √196

x = ±14


 (ii) 5x2= 625

Solution:

5x2 = 625

x2 = 625/5 = 125

x = √125 = ±5√5


 (iii) x2 + 1 = 101

Solution:

x2 = 101 – 1

x2 = 100

x = √100 = ±10


 (iv) 7x = 64/7x

Solution:

7x = 64/7x

49x2 = 64

x2 = 64/49

x = √(64/49)

x = ±8/7


 (v) (x + 8)2 – 5=31

Solution:

(x + 8)2 – 5 = 31

(x + 8)2 = 31 + 5

(x + 8)2 = 36

(x + 8) = √36

x + 8 = ±6

x + 8 = 6 or x + 8 = -6

x = 6 – 8 or x = -6 – 8

x = -2 or x = -14


 (vi) x^2/23/4 = 71/4

Solution:

x^2/23/4 = 71/4

x^2/23/4 = 29/4

2x2 – 3 = 29

2x2 = 29 + 3

2x2 = 32

x2 = 32/2 = 16

x = √16

x = ±4


 (vii) -4x2 + 324 = 0

Solution:

-4x2 + 324 = 0

-4x2 = -324

4x2 = 324

x2 = 324/4 = 81

x = √81 = ±9


 (viii) -37.5 x2 = -37.5

Solution:

-37.5 x2 = -37.5

x2 = -37.5/-37.5 = 1

x = √1 = ±1


3.In each of the following determine whether the given values of variables is a solution of the quadratic equation or not.

(i) x2 + 14x + 13 = 0; x = -1 ; x =-13

Solution:

If x = -1 then x2 + 14x + 13 = 0

x2 + 14x + 13 = (-1)2 + 14(-1) + 13

x2 + 14x + 13 = 1 – 14 + 13

x2 + 14x + 13 = 14 – 14

x2 + 14x + 13 = 0

Therefore x = -1 is the solution of the equation x2 + 14x + 13 = 0

Given, if x = -13 then x2 + 14x + 13

x2 + 14x + 13 = (-13)2 + 14(-13) + 13

x2 + 14x + 13 = 196  – 182 + 13

x2 + 14x + 13 = 27

if x = -13 then x2 + 14x + 13 ≠ 0

Therefore x = -13 is not the solution of the equation x2 + 14x + 13 = 0


 (ii) 7x2 – 12x = 0 ; x = 1/3

Solution:

Given, if x = 1/3 then 7x2 – 12x = 0

7x2 – 12x = 7(1/3)2 – 12(1/3)

7x2 – 12x = 7/912/3

7x2 – 12x = 7/9 – 4 = –29/9

If x = 1/3 then 7x2 – 12x ≠ 0

Therefore x = 1/3 is not the solution of the equation 7x2 – 12x = 0


 (iii) 2m2 – 6m + 3 = 0; m = 1/2

Solution:

Given, if m = 1/2 then 2m2 – 6m + 3 = 0

2m2 – 6m + 3 = 2(1/2)2 – 6(1/2) + 3

2m2 – 6m + 3 = 21/4 – 2 + 3

2m2 – 6m + 3 = 1/2

If x = 1/2 then 2m2 – 6m + 3 ≠ 0

Therefore x = 1/2 is not the solution of the equation 2m2 – 6m + 3 = 0


 (iv) y2 + 2y – 4 = 0 ; y = 2√2

Solution:

Given, if y = (2√2) then y2 + √2y – 4 = 0

y2 + √2y – 4 = (2√2)2 + √2(2√2)y – 4

y2 + √2y – 4 = 4×2 + 2×2 – 4

y2 + √2y – 4 = 8 + 4 – 4

y2 + √2y – 4 = 8

If y = 2√2 then y2 + √2y – 4 ≠ 0

Therefore y = (2√2) is not the solution of the equation y2 + √2y – 4 = 0


 (v) x/x+2 = 1/2 ; x = 2 and x = 1

Solution:

Given, if x = 2 then x/x+21/2 = 0

x/x+21/2 = 2/2 + 2  – 1/2 = 2/41/2 = 1/21/2 = 0

Therefore x = 2 is the solution of the equation x/x+21/2 = 0

Given, if x = 1 then x/x+21/2 = 0

x/x+21/2 = 1/1 + 2  – 1/2 = 1/31/2 = 2 – 3 /6  = -1/6

Therefore x = 1 is not the solution of the equation x/x+21/2 = 0


 (vi) 6x2 – x – 2 = 0 ; x = –1/2 and x = 2/3

Given, if x = (-1/2)  then  6x2 – x – 2 = 0

6x2 – x – 2 = 6(-1/2)2 – (-1/2) – 2

6x2 – x – 2 = 6(1/4) + 1/2 – 2

6x2 – x – 2 = 3/2 + 1/2 – 2

6x2 – x – 2 = 2 – 2

6x2 – x – 2 = 0

Therefore x = 2 is the solution of the equation  6x2 – x – 2 = 0


4.Solve the quadratic equations:

(i)If A= πr2, solve for r and find the value of r if A = 77 and π = 22/7

Solution:

A= πr2

77 = 22/7 x r2

77 x 7 = 22 r2

539 = 22r2

539/22 = r2

49/2 = r2

r = ±7/√2


 (ii) If r2 = l2 + d2 solve for d and find the value of d if r = 5 and l = 4

Solution:

r2 = l2 + d2

52 = 42 + d2

d2 = 52 – 42

d2 = 25 – 16 = 9

d = ±3


 (iii) If A = √3a^2/4 solve for v and find the value of a, if A = 16√3

Solution:

16√3 = √3a^2/4

16√3 x 4 = √3a2

a2 = 16√3 x 4/√3

a2 = 16 x 4 = 64

a = ±8


 (iv) if k = 1/2 mv2 solve for v and find the value of v, if k = 100 and m = 2

Solution:

k = 1/2 mv2

100 = 1/2 x 2 x v2

100 x 2 = 2v2

200/2 = v2

100 = v2

v = ±10


 (vi) If  v2 = u2 + 2as solve for v and find the value of v, if u = 0, a = 2, s = 100

Solution:

v2 = u2 + 2as

v2 = 0 + 2(2)(100)

v2 = 400

v2 = √400

v = ±20


Next  Exercise – Quadratic Equations – Exercise 9.3 – Class X


 

Quadratic Equations – Exercise 9.1 – Class X

Quadratic Equations – Exercise 9.1

  1. Check whether the following are quadratic equations:

(i) x2 – x = 0

(ii) x2 = 8

(iii) x2 + 1/2 x = 0

(iv) 3x – 10 = 0

(v) x229/4 x + 5 = 0

(vi) 5 – 6x = 2/5x2

(vii) √2x2 + 3x = 0

(viii) √3x = 22/13

(ix)x3 – 10x + 74 = 0

(x) x2 – y2 = 0

2.Simplify the following equations and check whether they are quadratic equations:

(i) x(x + 6) = 0

(ii) (x – 4)(2x – 3) = 0

(iii) (x + 2)(x – 7) = 0

(iv) (x + 1)2 = 2(x – 3)

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

(vi) (x + 2)3 = 2x(x2 – 1)

3.Represent the following in the form of quadratic equations:

(i) The product of two consecutive integers is 306

(ii) The length of a rectangular park (in meters) is one more than twice its breadth and its area is 528m2

(iii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/hr less, then it would have taken 3 hours more to cover the same distance.


Quadratic Equations – Exercise 9.1

  1. Check whether the following are quadratic equations:

(i) x2 – x = 0

Solution:

Given equation is x2 – x = 0

The equation is of the form ax2 + bx + c = 0, where c = 0

Therefore, x2 – x = 0 is quadratic equation.


 (ii) x2 = 8

Solution:

Given equation is x2 = 8

⇒ x2 + 0.x – 8 = 0

The equation is of the form ax2 + bx + c = 0, where coefficient of x is 0

Therefore, x2 – 8 = 0 is quadratic equation.


 (iii) x2 + 1/2 x = 0

Solution:

Given equation is x2 +1/2 x = 0

The equation is of the form ax2 + bx + c = 0, where c = 0

Therefore, x2 +1/2 x = 0 is quadratic equation.


 (iv) 3x – 10 = 0

Solution:

Given equation is 3x – 10 = 0

The equation is not of the form ax2 + bx + c = 0

Therefore,  3x – 10 = 0 is not a quadratic equation.


 (v) x229/4 x + 5 = 0

Solution:

Given equation is x229/4 x + 5 = 0

The equation is of the form ax2 + bx + c = 0.

Therefore, x229/4 x + 5 = 0 is quadratic equation.


 (vi) 5 – 6x = 2/5x2

Solution:

Given equation is 5  – 6x = 2/5x2

2/5x2 + 6x – 5 = 0

The equation is of the form ax2 + bx + c = 0.

Therefore, 5  – 6x = 2/5x2 is quadratic equation.


 (vii) √2x2 + 3x = 0

Solution:

Given equation is √2x2 + 3x = 0

The equation is of the form ax2 + bx + c = 0, where c = 0

Therefore, √2x2 + 3x = 0 is quadratic equation.


 (viii) √3x = 22/13

Solution:

Given equation is √3x = 22/13

⇒ √3x – 22/13 = 0

The equation is not of the form ax2 + bx + c = 0

Therefore, √3x = 22/13 is not a quadratic equation.


 (ix)x3 – 10x + 74 = 0

Solution:

Given equation is x3 – 10x + 74 = 0

The equation is not of the form ax2 + bx + c = 0

Therefore, x3 – 10x + 74 = 0 is not a quadratic equation.


 (x) x2 – y2 = 0

Solution:

Given equation is x2 – y2 = 0

The equation is of the form ax2 + bx + c = 0, where coefficient of x = 0

Therefore, x2 – y2 = 0 is a quadratic equation.


2.Simplify the following equations and check whether they are quadratic equations:

(i) x(x + 6) = 0

Solution:

x(x + 6) = 0

x2 + 6x = 0

The given equation is x2 + 6x = 0

The equation is of the form ax2 + bx + c = 0, where c = 0

Therefore, x(x + 6) = 0 is a quadratic equation.


 (ii) (x – 4)(2x – 3) = 0

Solution:

(x – 4)(2x – 3) = 0

2x2 – 3x – 8x + 12 = 0

2x2 – 11x + 12 = 0

Therefore, the given equation is (x – 4)(2x – 3) = 0

The equation is of the form ax2 + bx + c = 0

Therefore, (x – 4)(2x – 3) = 0 is a quadratic equation.


 (iii) (x + 2)(x – 7) = 0

Solution:

(x + 2)(x + 7) = 0

x2 + 7x + 2x + 14 = 0

x2 + 9x + 14 = 0

Therefore, the given equation is (x + 2)(x + 7) = 0

The equation is of the form ax2 + bx + c = 0

Therefore, (x + 2)(x + 7) = 0 is a quadratic equation.


 (iv) (x + 1)2 = 2(x – 3)

Solution:

(x + 1)2 = 2(x – 3)

x2 + 2x + 1 = 2x – 6

x2 + 2x  – 2x + 1 – 6 = 0

x2  – 5 = 0

Therefore, the given equation is (x + 1)2 = 2(x – 3)

The equation is of the form ax2 + bx + c = 0

Therefore, (x + 1)2 = 2(x – 3) is a quadratic equation.


 (v) (2x – 1)(x – 3) = (x + 5)(x – 1)

Solution:

(2x – 1)(x – 3) = (x + 5)(x – 1)

2x2 – 6x – x + 3 = x2 – x + 5x – 5

2x2 – 6x – x + 3 – x2 + x – 5x + 5 = 0

x2 – 12x + 8 = 0

Therefore, the given equation is (2x – 1)(x – 3) = (x + 5)(x – 1)

The equation is of the form ax2 + bx + c = 0

Therefore, (2x – 1)(x – 3) = (x + 5)(x – 1) is a quadratic equation.


 (vi) (x + 2)3 = 2x(x2 – 1)

Solution:

(x + 2)3 = 2x(x2 – 1)

(x + 2)(x2 + 2x + 4) = 2x3 – 2x

x3 + 2x2 + 4x + 2x2 + 4x + 8 – 2x3 – 2x = 0

-x3 + 4x2 + 6x + 8 = 0

Therefore, the given equation is (x + 2)3 = 2x(x2 – 1)

The equation is of the form ax2 + bx + c = 0

Therefore, (x + 2)3 = 2x(x2 – 1) is a quadratic equation.


3.Represent the following in the form of quadratic equations:

(i) The product of two consecutive integers is 306

(ii) The length of a rectangular park (in meters) is one more than twice its breadth and its area is 528m2

(iii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/hr less, then it would have taken 3 hours more to cover the same distance.

Solution:

 

 

(i) The product of two consecutive integers is 306

Let two consecutive numbers be x and x+1.

Then, x(x + 1) = 306

x2 + x – 306 = 0

(ii) The length of a rectangular park (in meters) is one more than twice its breadth and its area is 528m2

Area of rectangle = length x breadth

If length of the rectangular park = x m

then, breadth of the rectangular park = (2x + 1) m

(2x + 1)x = 528

2x2 + x – 528 = 0

(iii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/hr less, then it would have taken 3 hours more to cover the same distance.

Let the speed of train be x km/h

Time taken to travel 480 km = 480/x km/h

In second condition let the speed of the train = (x – 8) km/h

It is also given that the train will take 3 hours to cover the same distance.

Therefore, time taken to travel 480 km = (480/x + 3) km/h

Speed x time = distance

(x – 8)(480/x + 3) = 480

480 + 3x – 3840/x – 24 = 0

3x – 3840/x – 24 = 0

3x2 – 3840 – 24x = 0

x2 – 8x – 1280 = 0


Next Exercise – Quadratic equations – Exercise 9.2 – Class X


 

Polynomials – Exercise 8.5 – Class X

Previous Exercise – Polynomials – Exercise 8.4 – Class X


Polynomials – Exercise 8.5

  1. Find the quotients and remainder using synthetic division.

(i) (x3 + x2 – 3x + 5) /(x – 1)

(ii)  (3x3 – 2x2 + 7x  – 5) /(x + 3)

(iii) (4x3 – 16x2 – 9x – 36)/(x + 2)

(iv) (6x4 – 29x3 + 40x2  – 12)/(x – 3)

(v) (8x4 – 27x2 +  6x + 9)/(x + 1)

(vi) (3x3 – 4x2 –  10x + 6)/(3x – 2)

(vii) (8x4 – 27x2 + 6x – 5)/(4x + 1)

(viii) (2x4 – 7x3 – 13x2 + 63x – 48)/(2x – 1)

  1. If the quotient obtained on dividng (x4 + 10x3 + 35x2 + 50x + 29) by (x + 4) is (x3 – ax2 + bx + 6) then find a, b and also the remainder
  2. If the quotient obtained on dividing (8x4 – 2x2 + 6x – 7) by (2x +1) is (4x3 + px2 – qx + 3) then find p, q and also find the remainder.

Polynomials – Exercise 8.5 – Solutions:

  1. Find the quotients and remainder using synthetic division.

(i) (x3 + x2 – 3x + 5) /(x – 1)

Solution:

Polynomials – Exercise 8.5 – Class X

Therefore, q(x) = x2 + 2x – 1 , r(x) = 4


(ii)  (3x3 – 2x2 + 7x  – 5) /(x + 3)

Solution:

Polynomials – Exercise 8.5 – Class X

Therefore, q(x) = 3x2 – 11x + 40 and r(x) = -125


(iii) (4x3 – 16x2 – 9x – 36)/(x + 2)

Solution:

Polynomials – Exercise 8.5 – Class X

Therefore, q(x) =  4x2 – 24x + 39 and r(x) = -114


(iv) (6x4 – 29x3 + 40x2  – 12)/(x – 3)

Solution:

Polynomials – Exercise 8.5 – Class X

Therefore, q(x) = 6x3 – 11x2 + 7x + 21 and r(x) = 51


 (v) (8x4 – 27x2 +  6x + 9)/(x + 1)

Solution:

Polynomials – Exercise 8.5 – Class X

Therefore, q(x) = 8x3 – 8x2 – 19x + 25 and r(x) = -16


(vi) (3x3 – 4x2 –  10x + 6)/(3x – 2)

Solution:

Polynomials – Exercise 8.5 – Class X

Therefore, q(x) = 3x2  – 2x – 34/3 and r(x) = –14/9

⇒q(x) = x22/3 x – 34/9 and r(x) = –14/9


(vii) (8x4 – 2x2 + 6x – 5)/(4x + 1)

Solution:

Polynomials – Exercise 8.5 – Class X

Therefore, q(x) = 8x3  – 2x23/2x + 51/8 and r(x) = –211/32

⇒q(x) = 2x3  – 1/2x23/8 x + 51/32 and r(x) = –211/32


(viii) (2x4 – 7x3 – 13x2 + 63x – 48)/(2x – 1)

Solution:

Polynomials – Exercise 8.5 – Class X

Therefore, q(x) = 2x3  – 6x2 – 16x + 55 and r(x) = –41/2

⇒q(x) = x3  – 3x2 8x + 55/2 and r(x) = –41/2


  1. If the quotient obtained on dividing (x4 + 10x3 + 35x2 + 50x + 29) by (x + 4) is (x3 – ax2 + bx + 6) then find a, b and also the remainder

Solution:

Polynomials – Exercise 8.5 – Class X

Given, if the quotient obtained on dividing x4 + 10x3 + 35x2 + 50x + 29 by  (x + 4) is (x3 – ax2 + bx + 6).

q’(x) = x3 – ax2 + bx + 6

On dividing x4 + 10x3 + 35x2 + 50x + 29 by  (x + 4) quotient is x3 + 6x2 + 11x + 6 and the remainder is 5.

i.e., q(x) = x3 + 6x2 + 11x + 6 and r(x) = 5

q’(x) = q(x)

x3 + 6x2 + 11x + 6 = x3 – ax2 + bx + 6

Coefficient of x2 on both the sides, 6 = -a ⇒ a = -6

Coefficient of x on both the sides, 11 = b

Therefore, a = -6 and b = 11.

The remainder r(x) = 5.


  1. If the quotient obtained on dividing (8x4 – 2x2 + 6x – 7) by (2x +1) is (4x3 + px2 – qx + 3) then find p, q and also find the remainder.

Solution:

Given, if the quotient obtained on dividing (8x4 – 2x2 + 6x – 7) by (2x +1) is (4x3 + px2 – qx + 3).

q’(x) = (4x3 + px2 – qx + 3)

Polynomials – Exercise 8.5 – Class X

On dividing (8x4 – 2x2 + 6x – 7) by (2x +1) quotient is 8x3 – 4x2 + 6 and the remainder is -10

i.e., q(x) = 8x3 – 4x2 + 6 = 4x3 – 2x2 + 6  and r(x) = -10

q’(x) = q(x)

4x3 + px2 – qx + 3 = 4x3 – 2x2 + 6

Coefficient of x2 on both the sides, p = -2

Coefficient of x on both the sides, -q = 0 ⇒ q = 0

Therefore, p = -2 , q = 0 and r(x) = -10


 

Polynomials – Exercise 8.4 – Class X

Previous Exercise – Polynomials – Exercise 8.3 – Class X


Polynomials – Exercise 8.4

  1. In each of the following cases, use factor theorem to find whether g(x) is a factor of the polynomials p(x) or not.

(i) p(x) = x3 – 3x2 + 6x – 20 ; g(x) = x – 2

(ii) p(x) = 2x4 + x3 + 4x2 – x – 7 ; g(x) = x + 2

(iii) p(x) = 3x4 + 3x2 – 2x2 – 9x – 12 ; g(x) = x – 1/2

(iv) p(x) = 3x3 + x2 – 20x + 12 ; g(x) = 3x – 2

(v) p(x) = 2x4 + 3x3 – 2x2 – 9x – 12 ; g(x) = x2 – 3

  1. Find the valule of a if (x – 5) is a factor of (x3 – 3x2 + ax – 10)
  2. If (x3 + ax – bx + 10) is divisible by x2 – 3x + 2 , find the values of a and b
  3. If both (x – 2) and (x – 1/2)are factors of (ax2 + 5x + b), show that a = b

Polynomials – Exercise 8.4 –Solutions:

  1. In each of the following cases, use factor theorem to find whether g(x) is a factor of the polynomials p(x) or not.

(i) p(x) = x3 – 3x2 + 6x – 20 ; g(x) = x – 2

Solution:

Let p(x) = x3 – 3x2 + 6x – 20

By factor theorem, (x – 2) is a factor of p(x) if p(2) = 0

p(2) = 23 – 3(2)2 + 6(2) – 20

= 8 – 12 + 12 – 20

= – 12

⸫ (x – 2) is not a factor of p(x) = x3 – 3x2 + 6x – 20


(ii) p(x) = 2x4 + x3 + 4x2 – x – 7 ; g(x) = x + 2

Solution:

Let p(x) = 2x4 + x3 + 4x2 – x – 7

By factor theorem, (x + 2) is a factor of p(x) if p(-2) = 0

p(-2) = 2(-2)4 + (-2)3 + 4(-2)2 – (-2) – 7

= 32 – 8 + 16 + 2 – 7

= 35

⸫ (x + 2) is not a factor of p(x) = 2x4 + x3 + 4x2 – x – 7


 (iii) p(x) = 3x4 + 3x3 – 2x2 – 9x – 12 ; g(x) = x – 1/2

Solution:

Let p(x) = 3x4 + 3x3 – 2x2 – 9x – 12

By factor theorem, (x – 1/2) is a factor of p(x) if p(1/2) = 0

p(1/2) = 3(1/2)4 + 3(1/2)3 – 2(1/2)2 – 9(1/2) – 12

= 3(1/16) + 3(1/8) – 2(1/4) – 9(1/2) – 12

= – 263/16

⸫ (x – 1/2) is not a factor of p(x) = 3x4 + 3x2 – 2x2 – 9x – 12


 (iv) p(x) = 3x3 + x2 – 20x + 12 ; g(x) = 3x – 2

Solution:

Let p(x) = 3x3 + x2 – 20x + 12

By factor theorem, (3x – 2) is a factor of p(x) if p(2/3) = 0

p(2/3) = 3(2/3)3 + (2/3)2 – 20(2/3) + 12

= 0

⸫ (3x – 2) is a factor of p(x) = 3x3 + x2 – 20x + 12


 (v) p(x) = 2x4 + 3x3 – 2x2 – 9x – 12 ; g(x) = x2 – 3

Solution:

Let p(x) = 2x4 + 3x3 – 2x2 – 9x – 12

By factor theorem, (x2 – 3) is a factor of p(x) if p(√3) = 0

p (√3) = 2 (√3)4 + 3 (√3)3 – 2 (√3)2 – 9 (√3) – 12

= 2x3x3 + 3×3√3 – 2×3 – 9√3 – 12

= 18 + 9√3 – 6 – 9√3 – 12

= 0

⸫ (x2 – 3) is a factor of p(x) = 2x4 + 3x3 – 2x2 – 9x – 12


  1. Find the value of a if (x – 5) is a factor of (x3 – 3x2 + ax – 10)

Solution:

Let p(x) = x3 – 3x2 + ax – 10

By factor theorem, (x – 5) is a factor of p(x) if p(5) = 0

p(x) = x3 – 3x2 + ax – 10 = 0

53 – 3(52) + a(5) – 10 = 0

125 – 75 + 5a – 10 = 0

40 + 5a = 0

5a = – 40

a = –40/5 = – 8


  1. If (x3 + ax2 – bx + 10) is divisible by x2 – 3x + 2 , find the values of a and b

Solution:

p(x) = x3 + ax2 – bx + 10

g(x) = x2 – 3x + 2

g(x) = x2 –x – 2x + 2

g(x) = x(x – 1)-2(x – 1)

g(x) = (x – 1)(x – 2)

p(x) = (x3 + ax2 – bx + 10) is divisible by g(x) = (x – 1)(x – 2) , then we have p(2) = 0 and p(1) = 0

⇒ p(2) = p(1)

We have to find the values of a and b.

p(x) = x3 + ax2 – bx + 10

p(2) = (2)3 + a(2)2 – b(2) + 10 = 8 + 4a – 2b + 10 = 4a – 2b + 18 = 2a – b + 9 = 0

p(1) = (1)3 + a(1)2 – b(1) + 10 = 1 + a – b + 10 = a – b + 11 = 0

Since p(2) = p(1)

2a – b + 9 = a – b + 11

2a – a = 11 – 9

a = 2

Substitute the value of a in p(2) = 0

p(2) = a – b + 11 = 0

⇒ 2 – b + 11 = 0

-b = -11 – 2 = – 13

b = 13

Therefore, a = 2 and b = 13


  1. If both (x – 2) and (x – 1/2)are factors of (ax2 + 5x + b), show that a = b

Solution:

p(x) = ax2 + 5x + b

By factor theorem, (x – 2) and (x – 1/2) is a factor of p(x) then p(2) = p(1/2) = 0

p(2) = a(2)2 + 5(2) + b = 4a + b + 10 = 0

p(1/2) = a(1/2)2 + 5(1/2) + b = a/4 + 5/2 + b = 0

Since p(2) = p(1/2)

4a + b + 10 = a/4 + 5/2 + b

4a + 10 = a/4 + 5/2

4a – a/4 = 5/2 – 10

16a – a/4 = 5-20/2

15a/4 = 15/2

a/2 = 1

a = 2

Substitute the value of p(2) = 4a + b + 10 = 0

4a + b + 10 = 0

8 + b + 10 = 0

b = -18


Next exercise – Polynomials – Exercise 8.5 – Class X