NCERT10th MathematicsExercise 4.2Topic: 4 Quadratic Equations
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EXERCISE 4.2
Q1. Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0 (ii) 2x2 + x – 6 = 0
(iii) √2 x2 + 7 x + 5√2 = 0 (iv) 2x2 – x + 1/8 = 0
(v) 100x2 – 20x + 1 = 0
Explanation:
1) The quadratic equation is of the form ax2 + bx + c = 0, where a ≠ 0.
2) If α and β are the roots of the equation ax2 + bx + c = 0, we have,
aα2 + bα + c = 0 and aβ2 + bβ + c = 0. i.e. we say that,
α and β satisfy the equation ax2 + bx + c = 0.
3) We can solve this qardatic equation ax2 + bx + c = 0 using factorisation method.
Solution:
(i) x2 – 3x – 10 = 0
1) Quadratic equation: x2 – 3x – 10 = 0
here last term is 10 and its sign is minus, factorise 10 in such a way
that their difference will be - 3.
10 = (- 5) x (2) (- 5 + 2 = - 3).
x2 – 5x + 2x – 10 = 0x(x – 5) + 2(x – 5) = 0(x – 5)(x + 2) = 0
2) So, (x – 5) = 0 or (x + 2) = 0.
3) So, x = 5 or x = - 2.
(ii) 2x2 + x – 6 = 0
1) Quadratic equation: 2x2 + x – 6 = 0
here last term is 6 and its sign is minus, factorise 2 x 6 in such a way
that their difference will be 1.
2 x 6 = (3) x (- 4) (4 - 3 = 1).
2x2 – 3x + 4x – 6 = 0x(2x – 3) + 2(2x – 3) = 0(2x – 3)(x + 2) = 0
2) So, (2x – 3) = 0, or (x + 2) = 0
3) So, x = 3/2 or x = – 2.
(iii) √2x2 + 7x + 5√2 = 0
1) Quadratic equation: √2x2 + 7x + 5√2 = 0
here last term is 5√2 and its sign is plus, factorise √2 x 5√2 in such a way
that their sum will be 7.
√2 x 5√2 = 5 x 2 (5 + 2 = 7).
√2x2 + 5x + 2x + 5√2 = 0
x(√2x + 5) + √2(√2x + 5) = 0
(√2x + 5)(x + √2) = 0
2) So, (√2x + 5) = 0, or (x + √2) = 0
3) So, x = (- 5/√2) or x = - √2.
(iv) 2x2 – x + 1/8 = 0
1) Quadratic equation: 2x2 – x + 1/8 = 0
2) Multiply by 8, and we get: 16x2 – 8x + 1 = 0
here last term is 1 and its sign is plus, factorise 16 x 1 in such a way
that their sum will be – 8.
16 x 1 = (– 4) x (–4) (– 4 – 4 = – 8).
16x2 – 4x – 4x + 1 = 04x(4x – 1) – (4x – 1) = 0(4x – 1)(4x – 1) = 0
3) So, (4x – 1) = 0, or (4x – 1) = 0
4) So, x = 1/4 or x = 1/4.
(v) 100x2 – 20x + 1 = 0
1) Quadratic equation: 100x2 – 20x + 1 = 0
here last term is 1 and its sign is plus, factorise 100 x 1 in such a way
that their sum will be – 20.
100 x 1 = (– 10) x (– 10) (– 10 – 10 = – 20).
100x2 – 10x – 10x + 1 = 010x(10x – 1) – (10x – 1) = 0(10x – 1)(10x – 1) = 0
2) So, (10x – 1) = 0, or (10x – 1) = 0
3) So, x = 1/10 or x = 1/10.
Q2. Solve the problems given in Example 1.
(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.
Explanation:
1) The quadratic equation is of the form ax2 + bx + c = 0, where a ≠ 0.
2) Frame the quadratic equation from the given condition.
3) Solve this qardatic equation ax2 + bx + c = 0 using factorisation method.
Solution:
(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.
1) Let the number of marbles with John be x.
2) So, according to the problem, the number of marbles with Jivanti is (45 - x).
3) Both have lost 5 marbles, so now they have (x - 5) and (40 - x).
4) According to the problem,
(x - 5)(40 - x) = 124
x(40 - x) - 5(40 - x) = 124
40x - x2 - 200 + 5x = 124
- x2 + 40x + 5x - 200 - 124 = 0
- x2 + 45x - 324 = 0
x2 - 45x + 324 = 0
5) Quadratic equation: x2 – 45x + 324 = 0
here last term is 324 and its sign is plus, factorise 324 x 1 in such a way
that their sum will be – 45.
324 x 1 = (– 36) x (– 9) (– 36 – 9 = – 45).
x2 – 36x – 9x + 324 = 0x(x – 36) – 9(x – 36) = 0(x – 36)(x – 9) = 0
6) So, (x – 36) = 0, or (x – 9) = 0
7) So, x = 36 or x = 9.
8) John has 36 marbles and Jivanti has 9 marbles.
(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.
1) Let the number of toys produced in a day be x.
2) So, each toy's production cost on a day will be Rs (55 - x).
3) On one particular day, the total cost was Rs 750.
4) According to the problem,
x(55 - x) = 750
55x - x2 = 750
55x - x2 - 750 = 0
- x2 + 55x - 750 = 0
x2 - 55x + 750 = 0
5) Quadratic equation: x2 - 55x + 750 = 0
here last term is 750 and its sign is plus, factorise 750 x 1 in such a way
that their sum will be – 55.
750 x 1 = (– 25) x (– 30) (– 25 – 30 = – 55).
x2 – 25x – 30x + 750 = 0x(x – 25) – 30(x – 25) = 0(x – 25)(x – 30) = 0
6) So, (x – 25) = 0, or (x – 30) = 0
7) So, x = 25 or x = 30.
8) Hence, the number of toys produced that day is 25 or 30.
Q3. Find two numbers whose sum is 27 and whose product is 182.
1) Let the first number be x.
2) The second number will be (27 - x).
3) According to the problem,
x(27 - x) = 182
27x - x2 = 182
27x - x2 - 182 = 0
- x2 + 27x - 182 = 0
x2 - 27x + 182 = 0
4) Quadratic equation: x2 - 27x + 182 = 0
here last term is 182 and its sign is plus, factorise 182 x 1 in such a way
that their sum will be – 27.
182 x 1 = (– 14) x (– 13) (– 14 – 13 = – 27).
x2 – 14x – 13x + 182 = 0x(x – 14) – 13(x – 14) = 0(x – 14)(x – 13) = 0
5) So, (x – 14) = 0, or (x – 13) = 0
6) So, x = 14 or x = 14.
7) Hence, the numbers are 13 and 14.
Q4. Find two consecutive positive integers, the sum of whose squares is 365.
1) Let the first positive integer be x.
2) The second number will be (x + 1).
3) According to the problem,
x2 + (x + 1)2 = 365
x2 + x2 + 2x + 1 = 365
2x2 + 2x - 365 + 1 = 0
2x2 + 2x - 364 = 0
x2 + x - 182 = 0
4) Quadratic equation: x2 + x - 365 = 0
here last term is 182 and its sign is minus, factorise 182 x 1 in such a way
that their difference will be 1.
182 x 1 = (14) x (– 13) (14 – 13 = 1).
x2 + 14x – 13x – 182 = 0x(x + 14) – 13(x + 14) = 0(x + 14)(x – 13) = 0
6) So, (x + 14) = 0, or (x – 13) = 0
7) So, x = –14 or x = 13.
8) As our integer is positive, we have x = 13.
9) So the integers are 13 and 14.
Q 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.
1) Here, a right triangle's altitude depends on its base, so let the base be x.
2) So, altitude will be (x - 7).
3) According to the problem,
x2 + (x - 7)2 = 132
x2 + x2 - 14x + 49 = 132
2x2 - 14x + 49 = 169
2x2 - 14x + 49 - 169 = 0
2x2 - 14x - 120 = 0
x2 - 7x - 60 = 04) Quadratic equation: x2 - 7x - 60 = 0
here last term is 60 and its sign is minus, factorise 60 x 1 in such a way
that their difference will be - 7.
60 x 1 = (12) x (5) (5 – 12 = - 7).
x2 + 5x – 12x – 60 = 0x(x + 5) – 12(x + 5) = 0(x + 5)(x – 12) = 0
6) So, (x + 5) = 0, or (x – 12) = 0
7) So, x = – 5 or x = 12.
8) As the sides of a triangle can't be negative, ignore x = - 5, so we have x = 12.
9) So the base of a triangle is 12 and the altitude is 7.
Q6. 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 Rs 90, find the number of articles produced and the cost of each article.
1) Let the number of pottery articles produced in a day be x.
2) So, the cost of production of each article will be Rs (2x + 3).
3) On one particular day, the total cost was Rs 90.
4) According to the problem,
x(2x + 3) = 90
2x2 + 3x = 90
2x2 + 3x - 90 = 0
2x2 + 3x - 90 = 0
5) Quadratic equation: 2x2 + 3x - 90 = 0
here last term is 90 and its sign is minus, factorise 90 x 2 in such a way
that their difference will be 3.
90 x 2 = (15) x (– 12) (15 – 12 = 3).
2x2 + 15x – 12x + 90 = 0x(2x + 15) – 6(2x + 15) = 0(2x + 15)(x – 6) = 0
6) So, (2x + 15) = 0, or (x – 6) = 0
7) So, x = - 15/2 or x = 6.
8) As the number of articles produced can't be negative, so, x can't be - 15/2.
9) So the number of articles produced on that day is 6 and the cost of each article is
Rs 15.
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