Get detailed, step-by-step solutions for NCERT Class 10 Maths Chapter 4 Exercise 4.2 on Quadratic Equations.Class 10 Maths Exercise 4.2, Roots of Quadratic Equations, Factorization Method, Splitting the Middle Term, Quadratic Word Problems, Pythagoras Theorem Quadratic, Consecutive Integers Problem, Right Triangle Sides
Master finding the roots of quadratic equations using the Factorization Method (splitting the middle term) (Q.1). Solutions cover equations with fractions and irrational coefficients. Practice applying quadratic equations to solve problems involving consecutive integers, Pythagoras theorem (right triangles), and cost/production problems (Q.2-6). Essential for solving quadratic applications.
This exercise focuses on finding the roots of quadratic equations by the factorization method (splitting the middle term) and applying this technique to solve various word problems.
1. Finding Roots by Factorization
(i) $x^2 – 3x – 10 = 0$
We need two numbers whose product is $-10$ and sum is $-3$. These numbers are $-5$ and $2$.
$$x^2 – 5x + 2x – 10 = 0$$
$$x(x – 5) + 2(x – 5) = 0$$
$$(x – 5)(x + 2) = 0$$
The roots are $\mathbf{x = 5}$ and $\mathbf{x = -2}$.
(ii) $2x^2 + x – 6 = 0$
We need two numbers whose product is $2 \times (-6) = -12$ and sum is $1$. These numbers are $4$ and $-3$.
$$2x^2 + 4x – 3x – 6 = 0$$
$$2x(x + 2) – 3(x + 2) = 0$$
$$(2x – 3)(x + 2) = 0$$
The roots are $\mathbf{x = 3/2}$ and $\mathbf{x = -2}$.
(iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$
We need two numbers whose product is $\sqrt{2} \times 5\sqrt{2} = 5 \times 2 = 10$ and sum is $7$. These numbers are $5$ and $2$.
$$\sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0$$
$$x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0$$
$$(x + \sqrt{2})(\sqrt{2}x + 5) = 0$$
The roots are $\mathbf{x = -\sqrt{2}}$ and $\mathbf{x = -5/\sqrt{2}}$.
(iv) $2x^2 – x + \frac{1}{8} = 0$
Multiply the equation by 8 to clear the fraction:
$$16x^2 – 8x + 1 = 0$$
We need two numbers whose product is $16 \times 1 = 16$ and sum is $-8$. These numbers are $-4$ and $-4$.
$$16x^2 – 4x – 4x + 1 = 0$$
$$4x(4x – 1) – 1(4x – 1) = 0$$
$$(4x – 1)(4x – 1) = 0$$
The roots are $\mathbf{x = 1/4}$ and $\mathbf{x = 1/4}$.
(v) $100x^2 – 20x + 1 = 0$
We need two numbers whose product is $100 \times 1 = 100$ and sum is $-20$. These numbers are $-10$ and $-10$.
$$100x^2 – 10x – 10x + 1 = 0$$
$$10x(10x – 1) – 1(10x – 1) = 0$$
$$(10x – 1)(10x – 1) = 0$$
The roots are $\mathbf{x = 1/10}$ and $\mathbf{x = 1/10}$.
2. Solve the problems given in Example 1.
(Assuming Example 1 refers to the two problems formulated at the end of Exercise 4.1)
(i) Rectangular Plot Area: $2x^2 + x – 528 = 0$
Here $x$ is the breadth of the plot.
We need two numbers whose product is $2 \times (-528) = -1056$ and sum is $1$. These numbers are $33$ and $-32$.
$$2x^2 + 33x – 32x – 528 = 0$$
$$x(2x + 33) – 16(2x + 33) = 0$$
$$(x – 16)(2x + 33) = 0$$
Roots are $x = 16$ and $x = -33/2$.
Since length/breadth cannot be negative, we take $x = 16$.
- Breadth ($x$): $\mathbf{16}$ m
- Length ($2x+1$): $2(16) + 1 = \mathbf{33}$ m
(ii) Product of Two Consecutive Positive Integers: $x^2 + x – 306 = 0$
Here $x$ is the first integer.
We need two numbers whose product is $-306$ and sum is $1$. These numbers are $18$ and $-17$.
$$x^2 + 18x – 17x – 306 = 0$$
$$x(x + 18) – 17(x + 18) = 0$$
$$(x – 17)(x + 18) = 0$$
Roots are $x = 17$ and $x = -18$.
Since the integers must be positive, we take $x = 17$.
- First integer ($x$): $\mathbf{17}$
- Second integer ($x+1$): $17 + 1 = \mathbf{18}$
3. Find two numbers whose sum is 27 and product is 182.
Let the first number be $x$.
The second number is $27 – x$.
Product is 182:
$$x(27 – x) = 182$$
$$27x – x^2 = 182$$
$$x^2 – 27x + 182 = 0$$
We need two numbers whose product is $182$ and sum is $-27$. These numbers are $-13$ and $-14$.
$$x^2 – 13x – 14x + 182 = 0$$
$$x(x – 13) – 14(x – 13) = 0$$
$$(x – 13)(x – 14) = 0$$
Roots are $x = 13$ and $x = 14$.
If $x=13$, the other number is $27-13=14$. If $x=14$, the other number is $27-14=13$.
The two numbers are $\mathbf{13}$ and $\mathbf{14}$.
4. Find two consecutive positive integers, sum of whose squares is 365.
Let the first positive integer be $x$.
The next consecutive positive integer is $x + 1$.
Sum of their squares is 365:
$$x^2 + (x + 1)^2 = 365$$
$$x^2 + (x^2 + 2x + 1) = 365$$
$$2x^2 + 2x + 1 – 365 = 0$$
$$2x^2 + 2x – 364 = 0$$
Divide by 2:
$$x^2 + x – 182 = 0$$
We need two numbers whose product is $-182$ and sum is $1$. These numbers are $14$ and $-13$.
$$x^2 + 14x – 13x – 182 = 0$$
$$x(x + 14) – 13(x + 14) = 0$$
$$(x – 13)(x + 14) = 0$$
Roots are $x = 13$ and $x = -14$.
Since the integers must be positive, we take $x = 13$.
- First integer ($x$): $\mathbf{13}$
- Second integer ($x+1$): $13 + 1 = \mathbf{14}$
5. Right Triangle Sides
Let the base of the right triangle be $x$ cm.
The altitude is $7$ cm less than the base: $x – 7$ cm.
The hypotenuse is $13$ cm.
By the Pythagoras theorem (Base$^2$ + Altitude$^2$ = Hypotenuse$^2$):
$$x^2 + (x – 7)^2 = 13^2$$
$$x^2 + (x^2 – 14x + 49) = 169$$
$$2x^2 – 14x + 49 – 169 = 0$$
$$2x^2 – 14x – 120 = 0$$
Divide by 2:
$$x^2 – 7x – 60 = 0$$
We need two numbers whose product is $-60$ and sum is $-7$. These numbers are $-12$ and $5$.
$$x^2 – 12x + 5x – 60 = 0$$
$$x(x – 12) + 5(x – 12) = 0$$
$$(x – 12)(x + 5) = 0$$
Roots are $x = 12$ and $x = -5$.
Since the length of a side cannot be negative, we take $x = 12$.
- Base ($x$): $\mathbf{12}$ cm
- Altitude ($x-7$): $12 – 7 = \mathbf{5}$ cm
6. Cottage Industry Production Cost
Let the number of articles produced on that day be $x$.
The cost of production of each article is 3 more than twice the number of articles: $2x + 3$ (₹).
The total cost of production is $₹ 90$.
$$\text{Number of articles} \times \text{Cost per article} = \text{Total Cost}$$
$$x(2x + 3) = 90$$
$$2x^2 + 3x = 90$$
$$\mathbf{2x^2 + 3x – 90 = 0}$$
We need two numbers whose product is $2 \times (-90) = -180$ and sum is $3$. These numbers are $15$ and $-12$.
$$2x^2 + 15x – 12x – 90 = 0$$
$$x(2x + 15) – 6(2x + 15) = 0$$
$$(x – 6)(2x + 15) = 0$$
Roots are $x = 6$ and $x = -15/2$.
Since the number of articles cannot be negative, we take $x = 6$.
- Number of articles produced ($x$): $\mathbf{6}$
- Cost of each article ($2x + 3$): $2(6) + 3 = 12 + 3 = \mathbf{₹ 15}$