**KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.2** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.2.

## Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.2

Question 1.

Find the roots of the following quadratic equations by factorisation:

(i) x^{2} – 3x – 10 = 0 ,

(ii) 2x^{2} + x – 6 = 0

(iii) \(\sqrt{2}\)x^{2} + 7x + \(5 \sqrt{2}\) =0

(iv) 2x^{2} – x + \(\frac{1}{8}\) = 0

(v) 100x^{2} – 20x + 1 = 0

Solution:

(i) x^{2} – 3x – 10 = 0

x^{2} – 5x + 2x – 10 = 0

x(x – 5) + 2 (x – 5) = 0

(x – 5) (x + 2) = 0

If x + 2 = 0, then x = -2

If x – 5 = 0, then x = 5

∴ x = -2 OR +5.

(ii) 2x^{2} + x – 6 = 0

2x^{2} + 4x – 3x – 6 = 0

2x(x + 2) – 3(x + 2) = 0

(x + 2) (2x – 3) = 0

If x + 2 = 0, then x = -2

If 2x – 3 = 0, then 2x = 3 x = \(\frac{3}{2}\)

∴ x = -2 OR \(\frac{3}{2}\)

iv) 2x^{2} – x + \(\frac{1}{8}\) = 0 Multiply by 8

8(2x^{2} -x + \(\frac{1}{8}\)) = 0

16x^{2} – 8x + \(\frac{8}{8}\) = 0

16x^{2} – 8x + 1 = 0

16x^{2} – 4x – 4x + 1 = 0

4x (4x – 1) – 1 (4x – 1) =

(4x – 1) (4x – 1) = 0

4x – 1 = 0 or 4x – 1 = 0

4x = 1 or 4x = 1

x = \(\frac{1}{4}\) or x = \(\frac{1}{4}\)

x = \(\frac{1}{4}\) are the roots of the equation 2x^{2} – x + \(\frac{1}{8}\) = 0

(v) 100x^{2} – 20x + 1 = 0

100x^{2} – 10x – 10x + 1 = 0

10x(10x – 1) – 1 (10x – 1) = 0

(10x – 1) (10x – 1) = 0

(10x – 1)^{2} = 0

If 10x – 1 = 0, then

10x = 1

∴ x = \(\frac{1}{10}\)

∴ Two roots are \(\frac{1}{10} , \frac{1}{10}\)

Question 2.

Solve the problems:

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. We would like to find out the number of toys produced on that day.

Solution:

(i) Let John had x marbles and Jivanti had (45 – x) marbles.

When both of them lost 5 marbles then equation becomes (x – 5) × (45 – x – 5) = 124

⇒ (x – 5) × (40 – x) = 124

⇒ x^{2} – 45x + 324 = 0

⇒ x^{2} – 9x – 36x + 324 = 0

⇒ x(x – 9) – 36(x – 9) = 0

⇒ (x – 9)(x – 36) = 0

Either x – 9 = 0 or x – 36 = 0

Thus, x = 9 or x = 36

∴ If John had 9 marbles, then Jivanti had 45 – 9 = 36 marbles.

If John had 36 marbles, then Jivanti had 45 – 36 = 9 marbles.

(ii) Let the number of toys produced in a day be x.

Then cost of 1 toy = \(\frac{750}{x}\)

⇒ \(\frac{750}{x}\) = 55 – x

⇒ 750 = 55x – x^{2}

⇒ x^{2} – 55x + 750 = 0

⇒ x^{2} – 30x – 25x + 750 = 0

⇒ x(x – 30) – 25(x – 30) = 0

⇒ (x – 30)(x – 25) = 0

Either x – 30 = 0 or x – 25 = 0

x = 30 or x = 25

Question 3.

Find two numbers whose sum is 27 and the product is 182.

Solution:

Sum of two numbers be 27

Let one number be ‘x’ other number be ’27 – x’

Product of two numbers = 182

x(27 – x)= 182

27x – x^{2} = 182

x^{2} – 27x + 182 = 0

x^{2} – 14x – 13x+ 182 = 0

x (x – 14) – 13 (x – 14) = 0

(x – 14) (x – 13) = 0

x + 14 = 0 (or) x – 13 = 0

x = – 14 or x = 13

x = 14, 13

∴ Two numbers are 14, 13 (or) 13, 14

Question 4.

Find two consecutive positive integers, sum of whose squares is 365.

Solution:

In that numbers, let one of the numbers be ’x’.

It’s consecutive positive integer is (x + 1)

Sum of their squares is 365.

∴ (x)^{2} + (x + 1)^{2} = 365

x^{2} + x^{2} + 2x + 1 = 365

2x^{2} + 2x + 1 = 365

2x^{2} + 2x + 1 – 365 = 0

2x^{2} + 2x – 365 = 0

x^{2} + x – 182 = 0

x^{2} + 14x – 13x – 182 = 0

x(x + 14) – 13(x + 14) = 0

(x + 14) (x – 13) = 0

If x + 14 = 0, then x = -14

If x – 13 = 0, then x = 13

In these positive integer is 13.

∴ One number, x = 13

It consecutive number is, x + 1

= 13 + 1 = 14

∴ The Numbers are 14 and 13.

Question 5.

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm., find the other two sides.

Solution:

In ⊥∆ABC,

∠ABC = 90°.

AB is altitude,

BC is base AC is Hypotenuse

Let Base, BC = x cm.

if so considered, :

Altitude AB = (x – 7) cm.

Hypotenuse AC = 13 cm.

As per Pythagoras theorem,

In a right angled ABC,

AB^{2} + BC^{2} = AC^{2}

(x – 7)^{2} + (x)^{2} = (13)^{2}

x^{2} – 14x + 49 + x^{2}= 169

2x^{2} – 14x + 49 = 169

2x^{2} – 14x + 49 – 169 = 0

2x^{2} – 14x – 120 = 0

x^{2} – 7x – 60 = 0

x^{2} – 12x + 5x – 6 0 = 0

x(x – 12) + 5(x – 12) = o

(x – 12) (x + 5) =0

If x – 12 = 0, then x = 12

If x + 5 = 0, then x = -5

Positive value, x = 12

∴ Base, BC = x = 12 cm.

Altitude, AB = x – 7 = 12 – 7 = 5 cm.

Question 6.

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.

Solution:

Let the number of articles produced in a day = x

∴ Cost of production of each article = ₹ (2x + 3)

According to the condition,

Total cost = ₹ 90

⇒ x × (2x + 3) = 90

⇒ 2x^{2} + 3x = 90

⇒ 2x^{2} + 3x – 90 = 0

⇒ 2x^{2} – 12x + 15x – 90 = 0

⇒ 2x(x – 6) + 15(x – 6) = 0

⇒ (x – 6)(2x + 15) = 0

Either x – 6 = 0 or 2x + 15 = 0

⇒ x= 6 or x = \(\frac{-15}{2}\)

But the number of articles produced can never be negative.

⇒ x = \(\frac{-15}{2}\) is rejected

∴ Cost of production of each article = ₹ (2 × 6 + 3) = ₹ 15

Thus, the required number of articles produced is 6 and the cost of each article is ₹ 15.

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