# KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.2

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### Karnataka State Syllabus Class 10 Maths SolutionsChapter 10 Quadratic Equations Ex 10.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0 ,
(ii) 2x2 + x – 6 = 0
(iii) 2–√x2 + 7x + 52–√ =0
(iv) 2x2 – x + 18 = 0
(v) 100x2 – 20x + 1 = 0
Solution:
(i) x2 – 3x – 10 = 0
x2 – 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) 2x2 + x – 6 = 0
2x2 + 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 = 32
∴ x = -2 OR 32

iv) 2x2 – x + 18 = 0 Multiply by 8
8(2x2 -x + 18) = 0
16x2 – 8x + 88 = 0
16x2 – 8x + 1 = 0
16x2 – 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 = 14 or x = 14
x = 14 are the roots of the equation 2x2 – x + 18 = 0

(v) 100x2 – 20x + 1 = 0
100x2 – 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 = 110
∴ Two roots are 110,110

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
⇒ x2 – 45x + 324 = 0
⇒ x2 – 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 = 750x
⇒ 750x = 55 – x
⇒ 750 = 55x – x2
⇒ x2 – 55x + 750 = 0
⇒ x2 – 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 – x2 = 182
x2 – 27x + 182 = 0
x2 – 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
x2 + x2 + 2x + 1 = 365
2x2 + 2x + 1 = 365
2x2 + 2x + 1 – 365 = 0
2x2 + 2x – 365 = 0
x2 + x – 182 = 0 x2 + 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,
AB2 + BC2 = AC2
(x – 7)2 + (x)2 = (13)2
x2 – 14x + 49 + x2= 169
2x2 – 14x + 49 = 169
2x2 – 14x + 49 – 169 = 0
2x2 – 14x – 120 = 0
x2 – 7x – 60 = 0 x2 – 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
⇒ 2x2 + 3x = 90
⇒ 2x2 + 3x – 90 = 0
⇒ 2x2 – 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 = −152
But the number of articles produced can never be negative.
⇒ x = −152 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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