Quadratic Equation Exercise 9.8 – Class X

Quadratic Equations – Exercise 9.8 – Questions:

  1. Form the quadratic equation whose roots are
  2. i) 3, 5
  3. ii) 6, -5

iii) -3, 3/2

  1. iv) 2/3 , 3/2
  2. v) (2 + √3), (2 – √3)
  3. vi) (-3 + 2√5)(-3 – 2√5)

B.

  1. If m and n are the roots of the equation x2 – 6x + 2 = 0 find the value of

(i) (m + n)mn

(ii) 1/m + 1/n

(iii) m3n2 + n3m2

(iv) 1/n1/m

  1. IF a and b are the roots of the equation 3m2 = 6m + 5, find the value of

(i) a/b + b/a

(ii) (a + 2b)(2a + b)

  1. If p and q are the roots of the equation 2a2 – 4a + 1 = 0. Find the value of

(i) (p + q)2 + 4pq

(ii) p3 + q3

  1. Form a quadratic equation whose roots are p/q and q/p
  2. Find the value of ‘k’ so that the equation x2 + 4x + (k +2) = 0 has one root which is twice the other.
  3. Find the value of p so that the equation 2x2 – 3qx + 5q = 0 has one root which is twice the other.
  4. Find the value of p so that the equation 4x2 – 8px + 9 = 0 has roots whose difference is 4.
  5. If one root of the equation x2 + px + q = 0 is 3 times the other prove that 3p2 = 16q.

Quadratic Equations – Exercise 9.8 – Solution:

  1. Form the quadratic equation whose roots are

i) 3, 5

Solution:

Let m and n be the roots

Then m = 3 and n = 5

Sum of the roots = m + n = 3 + 5 = 8

Product of the roots = mn = 3 x 5 = 15

Standard form is x2 –(m+n)x + mn = 0

Thus, x2 – 8x + 15 = 0

 

ii) 6, -5

Solution:

Let m and n be the roots

Then m = 6 and n = -5

Sum of the roots = m + n = 6 – 5 = 1

Product of the roots = mn = 6 x (-5) = -30

Standard form is x2 –(m+n)x + mn = 0

Thus, x2 – 1x + (-30) = 0

x2 – x – 30 = 0

 

iii) -3, 3/2

Solution:

Let m and n be the roots

Then m = -3 and n = 3/2

Sum of the roots = m + n = -3 + 3/2 = -6+3/2 = –3/2

Product of the roots = mn = -3 x 3/2 = -9/2

Standard form is x2 –(m+n)x + mn = 0

Thus, x2 – (-3/2) x + (-9/2) = 0

x2 + 3/2x – 9/2 = 0

 

iv) 2/3 , 3/2

Solution:

Let m and n be the roots

Then m = 2/3 and n = 3/2

Sum of the roots = m + n = 2/3 + 3/2 = 4+9/6 = 13/6

Product of the roots = mn = 2/3 x 3/2 =1

Standard form is x2 –(m+n)x + mn = 0

Thus, x2 – (13/6) x + (1) = 0

x213/6x + 1 = 0

 

v) (2 + √3), (2 – √3)

Solution:

Let m and n be the roots

Then m = 2 + √3 and n = 2 – √3

Sum of the roots = m + n = (2 + √3)+( 2 – √3) = 4

Product of the roots = mn = (2 + √3)( 2 – √3) = 4 – 2√3 + 2√3 – 3 = 4 – 3 = 1

Standard form is x2 –(m+n)x + mn = 0

Thus, x2 – (4) x + (1) = 0

x2 – 4x + 1 = 0

 

vi) (-3 + 2√5)(-3 – 2√5)

Solution:

Let m and n be the roots

Then m = -3 + 2√5 and n = -3 – 2√5

Sum of the roots = m + n = (-3 + 2√5)+( -3 – 2√5) = -6

Product of the roots = mn = (-3 + 2√5)( -3 – 2√5) = 9 + 6√5 – 6√5 – 4×5 = 9 – 45 = – 36

Standard form is x2 –(m+n)x + mn = 0

Thus, x2 – (-6) x + (-36) = 0

x2 + 6x – 36 = 0


B.

  1. If m and n are the roots of the equation x2 – 6x + 2 = 0 find the value of

(i) (m + n)mn

(ii) 1/m + 1/n

(iii) m3n2 + n3m2

(iv) 1/n1/m

Solution:

The given quadratic equation is x2 – 6x + 2 = 0 which is of the form ax2 + bx + c = 0 where a = 1, b = -6 and c = 2

Quadratic Equation Exercise 9.8

Therefore, m = 3 + √7 and n = 3 – √7

(i) (m + n)mn

= (3+ √7+ 3 – √7)( 3 + √7)( 3 – √7)

= 6 (3 + √7)( 3 – √7)

=6 (9 – 3√7 + 3√7 – 7)

= 6 x 2

= 12

 

(ii) 1/m + 1/n

= 1/(3 + √7) + 1/(3 – √7)

= (3 – √7)+(3 + √7)/(3 – √7)(3+√7)

= 6/2

= 3

 

(iii) m3n2 + n3m2

= (3 + √7)3( 3 – √7)2 + (3 – √7)3( 3 + √7)2

= (90 + 34√7)(16 – 6√7) + (90 – 34√7)(16 + 6√7)

= 12 + 4√7 + 12 – 4√7

= 24

 

(iv) 1/n1/m

= 1/(3 – √7)1/(3 + √7)

= (3 + √7)-(3 – √7)/(3 – √7)(3+√7)

= 2√7/2

= √7


  1. If a and b are the roots of the equation 3m2 = 6m + 5, find the value of

(i) a/b + b/a

(ii) (a + 2b)(2a + b)

Solution:

The given equation is 3m2 = 6m + 5, this can be written as 3m2 – 6m – 5 = 0, which is of the form mx2 + nx + p = 0 where m = 3 ; n = -6 and  p = -5

Quadratic Equation Exercise 9.8

Quadratic Equation Exercise 9.8


  1. If p and q are the roots of the equation 2a2 – 4a + 1 = 0. Find the value of

(i) (p + q)2 + 4pq

(ii) p3 + q3

Solution:

Given p and q are the roots of the equation is 2a2 – 4a + 1 = 0, which is of the form ax2 + bx + c = 0 where a = 2, b = -4 , x = 1

Quadratic Equation Exercise 9.8

  1. Form a quadratic equation whose roots are p/q and q/p

Solution:

Let p/q and q/p be the roots of a quadratic equation

Sum of the roots = m + n = p/q + q/p = p^2+q^2/pq

Product of the roots = mn = p/q x q/p = 1

Standard form is x2 –(m+n)x + mn = 0

Thus, x2 – (p^2+q^2/pq) x + (1) = 0

pqx2 – (p2 + q2)x – pq = 0


  1. Find the value of ‘k’ so that the equation x2 + 4x + (k +2) = 0 has one root equal to zero.

Solution:

The given quadratic equation is x2 + 4x + (k + 2) = 0 where a = 1, b = 4 , c = k+2

The sum of the roots m+n = –b/a = – 4/1 = -4

The product of the roots mn = c/a = k+2/1 = k + 2

If one root is m then other root is zero.

Thus, m = m and n = 0

Therefore, m + n = – 4 ⇒ m + 0 = -4 ⇒m = -4

mn = k + 2 ⇒ 0 = k + 2 ⇒ k = -2


  1. Find the value of p so that the equation 2x2 – 3qx + 5q = 0 has one root which is twice the other.

Solution:

The given quadratic equation is 2x2  – 3qx + 5q = 0 where a = 2, b = -3q , c = 5q

The sum of the roots m+n = –(-3q)/2 = 3q/2

The product of the roots mn = c/a = 5q/2

If one root is m then other root is 2m.

Thus, m = m and n = 2m

Therefore, m + 2m = 3q/2 ⇒ 3m = 3q/2 ⇒m = q/2

mn = 5q/2 ⇒ 2m2 = (5q/2)

⇒ 2(q/2)2 = 5q/2

⇒ 2(q^2/4) = 5q/2

q^2/4 = 5q/4

⇒q2 = 5q

⇒q = 5


  1. Find the value of p so that the equation 4x2 – 8px + 9 = 0 has roots whose difference is 4.

Solution:

The given quadratic equation is 4x2 – 8px + 9 = 0 where a = 4, b = -8p , c = 9

The sum of the roots m+n = –(-8p)/4 = 2p

The product of the roots mn = c/a = 9/4

If one root is m then other root is (m – 4).

Thus, m = m and n = (m – 4)

m+n = 2p⇒ m+m-4 = 2p⇒ 2m = 2p + 4⇒m = p + 2

mn = 9/4 ⇒m(m – 4) = 9/4

⇒m2 – 4m – 9/4 = 0

⇒(p+2)2 – 4(p+2) – 9/4 = 0

⇒p2 + 4p + 4 – 4p – 8 – 9/4 = 0

⇒p225/4 = 0

⇒p2 = 25/4

⇒p = ±5/2


  1. If one root of the equation x2 + px + q = 0 is 3 times the other prove that 3p2 = 16q.

Solution:

The given quadratic equation is x2 + px + q = 0 where a = 1, b = p , c = q

The sum of the roots m+n = –(p)/1 = -p

The product of the roots mn = q/1 = q

If one root is m then other root is 3m.

Thus, m = m and n = 3m

m + n = -p ⇒ m+3m = -p

⇒4m = -p

⇒ m = -p/4

mn = q ⇒m(3m) = q

⇒3m2 = q

⇒3(-p/4)2 = q

⇒3p^2/16 = q

⇒3p2 = 16q