KC Sinha Mathematics Solution Class 10 Chapter 7 Quadratic Equations Exercise 7.2

Exercise 7.1
Exercise 7.2
Exercise 7.3
Exercise 7.4
Exercise 7.5

Exercise 7.2

Question 1 A

Check whether the following are quadratic equations:
(x − 2) (x + 1) = (x − 1) (x + 3)

Sol :
Given; (x − 2) (x + 1) = (x − 1) (x + 3)
⇒ x² + x − 2x − 2 = x² + 3x − x − 3
⇒ x² − x − 2 − x² − 2x + 3 = 0
⇒ −3x + 1 = 0
∵ The highest power of x in the equation is 1;
∴ It is not a quadratic equation.

Question 1 B

Check whether the following are quadratic equations:
(x − 2)² + 1 = 2x − 3

Sol :
Given; (x − 2)² + 1 = 2x − 3
⇒ x² − 2x + 4 + 1 = 2x − 3
⇒ x² − 2x + 5 − 2x + 3 = 0
⇒ x² − 4x + 8 = 0
∵ The highest power of x in the equation is 2;
∴ It is a quadratic equation

Question 1 C

Check whether the following are quadratic equations:
x (x + 1) + 8 = (x + 2) (x − 2)

Sol :
Given; x (x + 1) + 8 = (x + 2) (x − 2)
⇒ x² + x + 8 = x² − 2²
⇒ x² + x + 8 − x² + 4 = 0
⇒ x + 12 = 0
∵ The highest power of x in the equation is 1;
∴ It is not a quadratic equation.

Question 1 D

Check whether the following are quadratic equations:
(x − 3) (2x + 1) = x (x + 5)

Sol :
Given; (x − 3) (2x + 1) = x (x + 2)
⇒ 2x² + x − 6x − 3 = x² + 5x
⇒ 2x² + x − 6x − 3 − x² − 5x = 0
⇒ x² − 10x − 3 = 0
∵ The highest power of x in the equation is 2;
∴ It is a quadratic equation.

Question 1 E

Check whether the following are quadratic equations:
x (2x + 3) = x² + 1

Sol :
Given; x (2x + 3) = x² + 1
⇒ 2x² + 3x = x² + 1
⇒ 2x² + 3x − x² − 1 = 0
⇒ x² +3x − 1 = 0
∵ The highest power of x in the equation is 2;
∴ It is a quadratic equation.

Question 1 F

Check whether the following are quadratic equations:
x² + 3x + 1 = (x − 2)²

Sol :
Given; x² + 3x + 1 = (x − 2)²
⇒ x² + 3x + 1 = x² − 4x + 4
⇒ x² + 3x + 1 − x² − 4x − 4 = 0
⇒ −x − 3 = 0
∵ The highest power of x in the equation is 1;
∴ It is not a quadratic equation.

Question 1 G

Check whether the following are quadratic equations:
(x + 1) (x − 1) = (x + 2) (x + 3)

Sol :
Given; (x + 1) (x − 1) = (x + 2) (x + 3)
⇒ x² − 1² = x² + 5x + 6
⇒ x² − 1 − x² − 5x − 6 = 0
⇒ −5x − 7 = 0
∵ The highest power of x in the equation is 1;
∴ It is not a quadratic equation.

Question 1 H

Check whether the following are quadratic equations:

(x − 1)² = (x + 1)²
Sol :
Given; (x − 1)² = (x + 1)²
⇒ x² − x + 1² = x² + x + 1²
⇒ x² − x + 1 − x² − x − 1 = 0
⇒ −2x = 0
∵ The highest power of x in the equation is 1;
∴ It is not a quadratic equation.

Question 2 A

Check whether the following are quadratic equations:

(x + 2)³ = x³ − 4
Sol :
Given; (x + 2)³ = x³ − 4
⇒ x³ + 6x² + 12x + 2³ = x³ − 4
⇒ x³ + 6x² + 12x + 2³− x³ + 4 = 0
⇒ 6x² + 12x + 12 = 0
∵ The highest power of x in the equation is 2;
∴ It is a quadratic equation.

Question 2 B

Check whether the following are quadratic equations:

$x-\frac{1}{x}=8$
Sol :
Given; $x-\frac{1}{x}=8$
⇒ x² − 1 = 8x
⇒ x² − 8x − 1 = 0
∵ The highest power of x in the equation is 2;
∴ It is a quadratic equation.

Question 2 C

Check whether the following are quadratic equations :

$2 x^{2}-3 \sqrt{x}+5=0$
Sol :
Given; $2 x^{2}-3 \sqrt{x}+5=0$
⇒ $2 x^{2}+5=3 \sqrt{x}$
⇒ $\left(2 x^{2}+5\right)^{2}=(3 \sqrt{x})^{2}$
⇒ 4x⁴ + 20x² + 25 = 9x
⇒ 4x⁴ + 20x² − 9x + 25 = 0
∵ The highest power of x in the equation is 4;
∴ It is not a quadratic equation.

Question 2 D

Check whether the following are quadratic equations :

$x^{2}+\frac{1}{x}=5$
Sol :
Given; $x^{2}+\frac{1}{x}=5$
⇒ x³ + 1 = 5x
⇒ x³− 5x + 1 = 0
∵ The highest power of x in the equation is 3;
∴ It is not a quadratic equation.

Question 2 E

Check whether the following are quadratic equations :

$x^{2}-\frac{1}{x^{2}}=8$
Sol :
Given; $x^{2}-\frac{1}{x^{2}}=8$
⇒ x⁴− 1 = 8x²
⇒ x⁴− 8x²− 1 = 0
∵ The highest power of x in the equation is 4;
∴ It is not a quadratic equation.

Question 3 A

Represent the following situations mathematically:
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.

Sol :
Let ‘x’ be the number of marbles John had.
∴ Jivanti will have 45 − x marbles.
[∵ the total number of marbles is 45]
As both of them lost 5 marbles each; marbles they now have will be x − 5 and 40 − x respectively.
⇒ Product of the number of marbles = (x − 5) (40 − x)
∴ 40x − x² − 200 + 5x = 124 [∵ the product is 124]
⇒ x² − 45x + 324 = 0

Question 3 B

Represent the following situations mathematically:

A shopkeeper buys a number of books for Rs. 80. If he had bought four more books for the same amount, the book would have cost Re. 1 less.
Sol :
Let ‘x’ be the number of books bought by the shopkeeper.
∴ Cost of one book = Rs. 80 ÷ x.
[∵ the total cost of books is Rs. 80]
∴ Cost of one book when he buys 4 more books for same rate = Rs. 80 ÷ (x + 4).
When he buys four more books for the same amount; it will cost Re. 1 per book less than the previous.
∴ 80 ÷ (x + 4) = (80 ÷ x) − 1
⇒ $\frac{80}{x+4}=\frac{80}{x}-1$
⇒ 80x = (x + 4) (80 − x)
⇒ 80x = 80x − x² + 320 − 4x
⇒ x² + 4x + 320 = 0

Question 3 C

Represent the following situations mathematically:

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 Rs. 750. We would like to find out the number of toys produced on that day.
Sol :
Let ‘x’ be the number of toys produced on that day.
∴ Cost of production of each toy that day= Rs. 55 − x.
[∵ The cost of production of each toy (in rupees) is 55 minus the number of toys produced in a day]
∴ Cost of production of x toys = Rs. x (55 − x)
[∵ On a particular day, the total cost of production was Rs. 750.]
∴ x (55 − x) = 750
⇒ x² − 55x + 750 = 0

Question 3 D

Represent the following situations mathematically:

The sum of the squares of two positive integers is 117. If the square of the smaller number equals four times the larger number, we need to find the integers.
Sol :
Let ‘x’ and ‘y’ be the smaller and larger integer respectively.
∴ x² + y² = 117.
[∵ The sum of the squares of two positive integers is 117]
∴ x² = 4y
[∵ the square of the smaller number equals four times the larger number.]
∴ y² + 4y – 117 = 0