Understanding Quadratic Equations in the New NCERT Syllabus for Class 10
Quadratic equations form a fundamental concept in algebra, crucial for building a strong mathematical foundation. In the new Class 10 syllabus of NCERT, quadratic equations take center stage, preparing students to solve real-world problems efficiently using algebraic techniques.
EXERCISE 4.2
Explanation:
conditions:
aα2 + bα + c = 0 and aβ2 + bβ + c = 0. This means α and β are the solutions to the equation ax2 + bx + c = 0.
method, which involves breaking the middle term to find the roots.
Solution:
To factorize, note that the constant term is 10, and since its sign is negative, we need factors of 10 whose difference equals – 3.
The factors of 10 are (– 5) and (2), because – 5 + 2 = – 3.
x2 – 5x + 2x – 10 = 0x(x – 5) + 2(x – 5) = 0(x + 2)(x – 5) = 0
To factorize, note that the constant term is 6, and since its sign is negative, we need factors of 6 whose difference equals 1.
The factors of 6 are (– 3) and (4), because – 3 + 4 = 1.
2x2 – 3x + 4x – 6 = 0x(2x – 3) + 2(2x – 3) = 0(x + 2)(2x – 3) = 0
Here last term is 5√2 and its sign is positive. To factorize, we need to break down √2 x 5√2 in such a way
that their sum equals 7.
√2 x 5√2 = 5 x 2, and we need to find factors of 10 whose sum is 7. The factors of 10 are 5 and 2, because 5 + 2 = 7. Now we can use these factors to factorize the equation accordingly.
√2x2 + 5x + 2x + 5√2 = 0
x(√2x + 5) + √2(√2x + 5) = 0
(√2x + 5)(x + √2) = 0
The last term is 1, and its sign is positive. We need to factor 16 x 1 in such a way that their sum equals – 8. The factors of 16 that satisfy this condition are
16 x 1 = (– 4) x (–4) (– 4 – 4 = – 8).
16x2 – 4x – 4x + 1 = 04x(4x – 1) – (4x – 1) = 0(4x – 1)(4x – 1) = 0
The last term is 1 and its sign is positive. We need to factor 100 x 1 in such a way that their sum equals – 20. The factors of 100 x 1 = (– 10) x (– 10)
(– 10 – 10 = – 20).
100x2 – 10x – 10x + 1 = 010x(10x – 1) – (10x – 1) = 0(10x – 1)(10x – 1) = 0
(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:
Identify two numbers whose product is and whose sum is . Split the middle term using these numbers and factor by grouping. Solve the resulting linear factors to find the values of .
Solution:
marbles.
equation:
(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
The last term is 324 and its sign is positive. We need to factor 324 x 1 in such a way that their sum equals – 45. The factors of 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
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.
x(55 – x) = 750
55x – x2 = 750
55x – x2 – 750 = 0
– x2 + 55x – 750 = 0
x2 – 55x + 750 = 0
The last term is 750 and its sign is positive. We need to factor 750 x 1 in such a way that their sum equals – 55. The factors of 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
equation becomes:
x(27 – x) = 182
27x – x2 = 182
27x – x2 – 182 = 0
– x2 + 27x – 182 = 0
x2 – 27x + 182 = 0
The last term is 182 and its sign is positive. We need to factor 182 x 1 in such a way that their sum equals – 27. The factors of 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
equation:
x2 + (x + 1)2 = 365
x2 + x2 + 2x + 1 = 365
2x2 + 2x – 365 + 1 = 0
2x2 + 2x – 364 = 0
x2 + x – 182 = 0
The last term is 182 and its sign is negative. We need to factor 182 x 1 in such a way that the difference between the factors is 1. The factors of
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
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.
The last term is 60 and its sign is negative. We need to factor 60 x 1 in such a way that the difference between the factors is – 7. The factors of
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
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.
x(2x + 3) = 90
2x2 + 3x = 90
2x2 + 3x – 90 = 0
2x2 + 3x – 90 = 0
The last term is 90 and its sign is negative. We need to factor 90 x 1 in such a way that the difference between the factors is 3. The factors of
90 x 1 = (15) x (– 12)
(15 – 12 = 3).
2x2 + 15x – 12x + 90 = 0x(2x + 15) – 6(2x + 15) = 0(2x + 15)(x – 6) = 0
article is Rs 15.