## Quadratic Equations Exercise 9.11 – Questions

- Find two consecutive positive odd numbers such that the sum of their squares is equal to 130.
- Find the whole number such that four times the number subtracted from three times the square of the number makes 15.
- The sum of two natural numbers is 8. Determine the numbers, if the sum of their reciprocals is
^{8}/_{15} - A two digit number is such that the product of the digits is 12. When 36 is added to this number the digits interchange their places. Determine the number
- Find three consecutive positive integers such that the sum of the square of the first and the product of other two is 154.
- The ages of Kavya and Karthik are 11 years and 14 years. In how many years times will the product of their ages be 304?
- The age of the mother is twice the square of the age of her son. Eight years hence, the age of the mother will be 4 years more than three times the age of her son. Find their present age.
- The area of a rectangle is 56 cm
^{2}. If the measure of its base is represented by (x + 5) and the measure of its height by (x – 5), find the dimensions of the rectangle. - The altitude of a triangle is 6 cm greater than its base. If its area is 108cm
^{2} . Find its base and height. - In rhombus ABCD, the diagonals AC and BD intersect at E. If AE = x, BE = (x + 7) and AB = (x + 8), find the lengths of the diagonals AC and BD.
- In an isosceles triangle ABC, AB = BC and BD is the altitude to base AC.IF DC = x , BD = 2x – 1 and BC = 2x + 1, fibd the lengths of all three sides of the triangle.
- A motor boat whose speed is 15km/hr in still water 30 km downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of the stream.
- A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.
- Shwetha takes 6 ways less than the number of days taken by Ankitha to complete a piece of work. If both Shwetha and Ankitha together can complete the same work in 4 days, in how many days will Ankitha alone complete the work?

## Quadratic Equations Exercise 9.11 – Solutions:

** Find two consecutive positive odd numbers such that the sum of their squares is equal to 130.**

Solution:

Let one of the odd positive number be x.

The other odd positive number be (x + 2).

The sum of the numbers = x^{2} + (x + 2)^{2} = 130

x^{2} + x^{2} + 2x + 4 = 130

2x^{2} + 2x + 4 = 130

2x^{2} + 2x – 126 = 0

x^{2} + x – 63 = 0

x^{2} + 9x – 7x – 63 = 0

x(x + 9) -7(x + 9) = 0

(x – 7)(x + 9) = 0

x = 7 or x = -9

Therefore, one of the odd positive number, x = 7 and the other odd positive number is x + 2 = 7 + 2 = 9

Therefore, two consecutive positive odd numbers 7 and 9.

** Find the whole number such that four times the number subtracted from three times the square of the number makes 15.**

Solution:

Let the whole number be x.

Four times the number be 4x and three times the number be 3x^{2}

Therefore,

4x – 3x^{2} + 15 = 0

-3x^{2} + 4x + 15 = 0

-3x^{2} + 9x – 5x + 15 = 0

3x(x – 3)-5(x – 3) = 0

(x – 3)(3x – 5) = 0

x = 3 or 3x = 5

x = 3 or x = ^{5}/_{3}

Therefore the whole number is 3.

** The sum of two natural numbers is 8. Determine the numbers, if the sum of their reciprocals is **^{8}/_{15}

Solution:

The sum of two natural numbers is 8. i.e., x + y = 8

x = 8 – y

The sum of their reciprocals is ^{8}/_{15} i.e., ^{1}/_{x} + ^{1}/_{y} = ^{8}/_{15}

^{1}/_{x} + ^{1}/_{y} = ^{8}/_{15}

^{y + x}/_{xy} = ^{8}/_{15}

(x + y)15 = 8xy

15 x 8 = 8 xy

xy = 15

(8 – y)y = 15

8y – y^{2} = 15

-y^{2} + 8y – 15 = 0

-y^{2} + 3y + 5y – 15 = 0

-y(y – 3)+5(y – 3) = 0

(y – 3)(-y + 5) = 0

y = 3 or –y + 5 = 0

y = 3 or y = 5

Therefore the numbers be 3, 5 .

** A two digit number is such that the product of the digits is 12. When 36 is added to this number the digits interchange their places. Determine the number**

Solution:

Let ten’s digit of the number = x and its unit digit = y. Then two digit number is 10x + y

xy = 12 and 10x + y + 36 = 10y + x

y = ^{12}/_{x} and 9x + 36 = 9y i.e., x + 4 = y

Substituting y = ^{12}/_{x} in x + 4 = 4; we get,

x + 4 = ^{12}/_{x}

x^{2} + 4x – 12 = 0

x^{2} +6x – 2x – 12 = 0

x(x + 6)-2(x + 6) = 0

(x + 6)(x – 2) = 0

x = -6 and x = 2

Since x is a digit, therefore, x = 2 and y = ^{12}/_{x} = ^{12}/_{2} = 6

Therefore the required number is 10x + y = 10×2 + 6 = 26

** Find three consecutive positive integers such that the sum of the square of the first and the product of other two is 154.**

Solution:

Let three consecutive numbers be x, x + 1, x + 2

x^{2} +(x+1)(x+2) = 154

x^{2} + (x^{2} + 2x + x + 2) = 154

2x^{2} + 3x + 2 = 154

2x^{2} + 3x – 152 = 0

2x^{2} + 19x – 16x – 152 = 0

x(2x + 19) – 8(2x +19) = 0

(2x + 19)(x – 8) = 0

2x + 19 = 0 or x – 8 = 0

2x = -19 or x = 8

x = –^{19}/_{2} or x = 8

Therefore, three consecutive numbers be 8, 9 and 10.

** The ages of Kavya and Karthik are 11 years and 14 years. In how many years time will the product of their ages be 304?**

Solution:

The age of Kavya be 11 years and the age of the Karthik be 14 years.

In x years the product of their ages will be 304. i.e., (x + 11)(x + 14) = 304

x^{2} + 14x + 11x + 154 = 304

x^{2} + 25x + 154 – 304 = 0

x^{2} + 25x + 150 = 0

x^{2} + 30x – 5x + 150 = 0

x(x + 30) – 5(x + 30) = 0

(x + 30)(x – 5) = 0

x + 30 = 0 or x – 5 = 0

x = -30 or x = 5

Therefore, in 5 years the product of their ages will be 304.

** The age of the mother is twice the square of the age of her son. Eight years hence, the age of the mother will be 4 years more than three times the age of her son. Find their present age.**

Solution:

Let the age of the mother be x.

Hence the present age of son = 2x^{2}

Given the age of the mother is twice the square of the age of her son i.e., x = 2x^{2}

Age of son after 8 years = (x + 8)years

Age of mother after 8 years = (2x^{2} + 8) years

Given that, (2x^{2} + 8) = 3(x + 8) + 4

2x^{2} + 8 = 3x + 24 + 4

2x^{2} – 3x + 8 – 28 = 0

2x^{2} – 3x – 20 =0

2x^{2} – 8x + 5x – 20 = 0

2x(x – 4) +5(x – 4) = 0

(2x + 5)(x – 4) = 0

x – 4 = 0 or 2x + 5 = 0

x = 4 or 2x = -5

x = 4 or x = –^{5}/_{2}

Thus, x = 4

Hence present age of son is 4 years and present age of mother is 2x^{2} = 2(4)^{2} = 32.

** The area of a rectangle is 56 cm**^{2}. If the measure of its base is represented by (x + 5) and the measure of its height by (x – 5), find the dimensions of the rectangle.

Solution:

A = 56 m^{2}

56 = (x + 5)(x – 5)

56 = x^{2} – 5x + 5x – 25

56 = x^{2} – 25

56 + 25 = x^{2}

81 = x^{2}

x = √81 = 9

Therefore, base of the rectangle = x + 5 = 9 + 5 = 14

height of the rectangle = x – 5 = 9 – 5 = 4

** The altitude of a triangle is 6 cm greater than its base. If its area is 108cm**^{2}. Find its base and height.

Solution:

Let the base of the triangle be x.

Then altitude of the triangle = x + 6

Area of the triangle = 108 cm^{2}

Formula to find the area of triangle = ^{1}/_{2} x base x height

108 = ^{1}/_{2} x(x + 6)

216 = x(x + 6)

216 = x^{2} + 6x

x^{2} + 6x – 216 = 0

x^{2} + 18x – 12x – 216 = 0

x(x + 18) -12(x + 18) = 0

(x – 12)(x + 18) = 0

x – 12 = 0 or x + 18 = 0

x = 12

Let the base of the triangle be 12cm

Then height of the triangle 18cm

** In rhombus ABCD, the diagonals AC and BD intersect at E. If AE = x, BE = (x + 7) and AB = (x + 8), find the lengths of the diagonals AC and BD.**

Solution:

. In rhombus ABCD, the diagonals AC and BD intersect at E. Therefore, angles at E are right angles. Hence, all the triangles formed by the diagonals AC and BD are right angled triangle.

In right angled triangle ABE, by Pythagoras theorem. We have

AB^{2} = AE^{2} + EB^{2}

(x + 8)^{2} = x^{2} + (x + 7)^{2}

x^{2} + 16x + 64 = x^{2} + x^{2} + 14x + 49

x^{2} + 16x + 64 – x^{2} – x^{2} – 14x – 49 = 0

-x^{2} + 2x + 15 = 0

-x^{2} + 5x – 3x + 15 = 0

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

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

x + 3 = 0 or –x +5 = 0

x = -3 or x = 5

The diagonal AC = AE + EC = 2(AE) = 2x = 10 cm

The diagonal BD = BE + ED = 2(BE)= 2(x+7) = 2(5 +7) = 2(12) = 24 cm

** **

** In an isosceles triangle ABC, AB = BC and BD is the altitude to base AC.IF DC = x , BD = 2x – 1 and BC = 2x + 1, find the lengths of all three sides of the triangle.**

Solution:

Given, AB = BC = 2x + 1

In an isosceles triangle ABC, AB = BC and BD is the altitude to base AC.

Therefore triangle BCD is a right angled triangle.

By Pythagoras theorem, BC^{2} = BD^{2} + CD^{2}

(2x + 1)^{2} = (2x – 1)^{2} + x^{2}

4x^{2} + 4x + 1 = 4x^{2} – 4x + 1 + x^{2}

4x^{2} + 4x + 1 – 4x^{2} + 4x – 1 – x^{2} = 0

-x^{2} + 8x = 0

x(-x + 8) = 0

-x + 8 = 0

x = 8

We know, AB = BC = 2x + 1 = 2(8) + 1 = 17cm

AC = AD + DC = 2(DC) = 2x = 2(8) = 16 cm

** A motor boat whose speed is 15km/hr in still water 30 km downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of the stream.**

Solution:

Let the speed of the stream be x km/hr. Then speed downstream = (15 + x) kn/hr

Speed upstream = (15 –x )km/hr

given, it took 4 hours 30 minutes to travel back to same place.

So, we have,

^{30}/_{15 + x} – ^{30}/_{15 – x} = 4^{1}/_{2}

^{30*30}/_{(15+x)(15-x)} = 4^{1}/_{2}

^{900}/_{225 – x^2} = ^{9}/_{2}

900 *2 = (225 – x^{2})*9

1800 = 2025 – 9x^{2}

9x^{2} = – 1800 + 2025

9x^{2} = 225

x^{2} = ^{225}/_{9}

x^{2} = 25

x = 5km/hr

** A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.**

Solution:

Let x be the cost price of the article.

The selling price of the article Rs. 24

x = ^{24 – x}/_{x} * 100

x^{2} = (24 – x)100

x^{2} = 2400 – 100 x

x^{2} + 100x – 2400 = 0

x^{2} + 120x – 20x – 2400 = 0

x(x + 120) -20(x + 120) = 0

(x + 120)(x – 20) = 0

x = -120 or x = 20

Therefore, cost price of the article is 20 Rs.

** Shwetha takes 6 ways less than the number of days taken by Ankitha to complete a piece of work. If both Shwetha and Ankitha together can complete the same work in 4 days, in how many days will Ankitha alone complete the work?**

Solution:

Let Shwetha take x – 6 days and Ankitha takes x days.

Shwetha one day work = ^{1}/_{(x – 6) }and Ankitha one day work = ^{1}/_{x}

Given, (A + B) can finish work in 4 day.

^{1}/_{x} + ^{1}/_{(x – 6) } = ^{1}/_{4}

^{(x-6)+x}/_{x(x-6)} = ^{1}/_{4}

^{2x -6}/_{x(x – 6) }= ^{1}/_{4}

4(2x – 6) = x(x – 6)

8x – 24 = x^{2} – 6x

x^{2} – 6x – 8x + 24 = 0

x^{2} – 14x + 24 = 0

x^{2} – 12x – 2x + 24 = 0

x(x – 12)-2(x – 12) = 0

(x – 12)(x – 2) = 0

x = 12 days taken by Ankitha alone.