Class ten NCERT solutions Quadratic Equations
Chapter 4: Quadratic Equations
TEXTBOOK’S EXERCISE 4.2
1. Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0 (ii) 2x2 + x – 6 = 0
(v) 100x2 – 20x + 1 = 0
Sol. (i) x2 – 3x – 10 = 0 ⇒ x2 – 5x + 2x – 10 = 0
⇒ x(x – 5) + 2(x – 5) = 0
⇒ (x – 5) (x + 2) = 0
Therefore, x – 5 = 0 or x + 2 = 0
i.e., x = 5 or x = –2
Hence, 5 and –2 are the roots of the given quadratic equation.
(ii) 2x2 + x – 6 = 0
⇒ 2x2 + 4x – 3x – 6 = 0
⇒ 2x(x + 2) – 3(x + 2) = 0
⇒ (x + 2) (2x – 3) = 0
⇒ x + 2 = 0 or 2x – 3 = 0,
⇒ x = –2 or x = 3/2
Hence, –2 and 3/2 are the roots of the given quadratic equation.
are the roots of the given quadratic equation.
Hence, 1/4 and 1/4 are the roots of the given quadratic equation.
(v) 100x2 – 20x + 1 = 0
⇒ 100x2 – 10x – 10x + 1 = 0
⇒ 10x(10x – 1) – 1(10x – 1) = 0
⇒ (10x – 1) (10x – 1) = 0
⇒ x = 1/ 10 or x = 1/ 10
Hence, 1 /10 and 1/ 10 are the roots of given quadratic equation.
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 Rs 750. We would like to find out the number of toys produced on that day.
Sol. (i) Let the number of marbles John had be x.
A number of marbles Jivanti had = 45 – x.
The number of marbles left with John, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles
= 45 – x – 5 = 40 – x
Then, required product = (x – 5) (40 – x)
= 40x – x2 – 200 + 5x
= –x2 + 45x – 200
As per condition: –x2 + 45x – 200 = 124
⇒ –x2 + 45x – 324 = 0
⇒ x2 – 45x + 324 = 0
⇒ x2 – 9x – 36x + 324 = 0
⇒ x(x – 9) – 36 (x – 9) = 0
⇒ (x – 36) (x – 9) = 0 ⇒ x = 36 or x = 9
Hence, they had started with 36 and 9 marbles.
Class ten NCERT solutions Quadratic Equations
(ii) Let the number of toys produced on a particular day be x.
The cost of production of each toy on that day = Rs (55 – x)
The total cost of production on that day = Rs x(55 – x)
Therefore, x(55 – x) = 750
⇒ 55x – x2 = 750
⇒ x2 – 55x + 750 = 0
⇒ x2 – 25x – 30x + 750 = 0 ⇒ x(x – 25) – 30(x – 25) = 0
⇒ (x – 25)(x – 30) = 0
⇒ x = 25 or x = 30
Hence, the number of toys produced on a particular day can be 25 or 30.
3. Find two numbers whose sum is 27 and product is 182.
Sol. Let first number = x.
And second number = 27 – x
Their product = x(27 – x)
As per condition: x(27 – x) = 182
⇒ 27x – x2 = 182
⇒ –x2 + 27x – 182 = 0
⇒ x2 – 27x + 182 = 0
⇒ x2 – 13x – 14x + 182 = 0
⇒ x(x – 13) – 14(x – 13) = 0
⇒ (x – 13) (x – 14) = 0
⇒ x = 13 or x = 14
Hence, the required numbers are 13 and 14.
Class ten NCERT solutions Quadratic Equations
4. Find two consecutive positive integers, the sum of whose squares is 365.
Sol. Let first integer = x
Second consecutive integer = x + 1
Sum of their squares = x2 + (x + 1)2
= x2 + x2 + 1 + 2x = 2x2 + 2x + 1
As per condition: 2x2 + 2x + 1 = 365
⇒ 2x2 + 2x = 364 ⇒ x2 + x = 182
⇒ x2 + x – 182 = 0
⇒ x2 + 14x – 13x – 182 = 0
⇒ x(x + 14) – 13 (x + 14) = 0
⇒ (x + 14) (x – 13) = 0
⇒ x = 13 or x = –14
Since, x is a positive integer, therefore x _ −14.
If x = 13 then, 2nd number = 13 + 1 = 14
Hence, the required two positive consecutive integers are 13 and 14.
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.
Sol. Let base of the right triangle = x cm
Height of the right triangle = (x – 7) cm
By Pythagoras theorem,
(Hyp.)2 = (Base)2 + (Height)2
⇒ (13)2 = x2 + (x – 7)2
⇒ 169 = x2 + x2 + 49 – 14x
⇒ 2x2 – 14x – 120 = 0
⇒ x2 – 7x – 60 = 0
⇒ x2 – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12) (x + 5) = 0
⇒ x = 12 or x = –5
∵ Side cannot be negative, so, x = 12 cm
Now, x – 7 = 12 – 7 = 5 cm.
Hence, the length of the other two sides is 5 cm and 12 cm.
Class ten NCERT solutions Quadratic Equations
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.
Sol. Let the number of articles produced = x
Cost of production of an article = Rs (2x + 3)
Total cost of production = Number of articles produced on that day × Cost of production of each article
As per condition : x(2x + 3) = 90
⇒ 2x2 + 3x – 90 = 0
⇒ 2x2 + 15x – 12x – 90 = 0
⇒ x (2x + 15) –6(2x + 15) = 0
⇒ (2x + 15) (x – 6) = 0 ⇒ x = 6 or x = − 15/2
As, number of articles cannot be negative. Therefore, x ≠ − 15/ 2
Hence, number of articles produced on that day = 6
So, cost of each article = 2x + 3
= Rs (2 × 6 + 3) = Rs 15