Quadratic Equations – Exercise 4.2

Question 1

Find the roots of the following quadratic equations by factorisation:

(i) $x^2 - 3x - 10 = 0$

(ii) $2x^2 + x - 6 = 0$

(iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$

(iv) $2x^2 - x + \frac{1}{8} = 0$

(v) $100x^2 - 20x + 1 = 0$

(i)
$x^2 - 3x - 10 = 0$
Split middle term: $x^2 - 5x + 2x - 10 = 0$
$x(x - 5) + 2(x - 5) = 0 \Rightarrow (x - 5)(x + 2) = 0$.
✅ Roots are $5$ and $-2$.

(ii)
$2x^2 + x - 6 = 0$
Split middle term: $2x^2 + 4x - 3x - 6 = 0$
$2x(x + 2) - 3(x + 2) = 0 \Rightarrow (2x - 3)(x + 2) = 0$.
✅ Roots are $\frac{3}{2}$ and $-2$.

(iii)
$\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$
Product = $5\sqrt{2} \times \sqrt{2} = 10$. Sum = 7. Numbers are 5 and 2.
$\sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0$
$x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0$
$(\sqrt{2}x + 5)(x + \sqrt{2}) = 0$.
✅ Roots are $-\frac{5}{\sqrt{2}}$ and $-\sqrt{2}$.

(iv)
$2x^2 - x + \frac{1}{8} = 0 \Rightarrow 16x^2 - 8x + 1 = 0$
$(4x - 1)^2 = 0$.
✅ Roots are $\frac{1}{4}, \frac{1}{4}$.

(v)
$100x^2 - 20x + 1 = 0$
$(10x - 1)^2 = 0$.
✅ Roots are $\frac{1}{10}, \frac{1}{10}$.

Word Problems

Solve the following problems:

(i) Find two numbers whose sum is 27 and product is 182.

(ii) Find two consecutive positive integers, sum of whose squares is 365.

(iii) The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

(iv) 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.

(i) Let numbers be $x$ and $27 - x$.
$x(27 - x) = 182 \Rightarrow 27x - x^2 = 182 \Rightarrow x^2 - 27x + 182 = 0$.
Factors of 182 summing to 27 are 13 and 14.
$(x - 13)(x - 14) = 0$.
✅ The numbers are 13 and 14.

(ii) Let integers be $x$ and $x + 1$.
$x^2 + (x + 1)^2 = 365 \Rightarrow 2x^2 + 2x + 1 = 365$
$2x^2 + 2x - 364 = 0 \Rightarrow x^2 + x - 182 = 0$.
Factors of 182 with difference 1 are 14 and 13.
$(x + 14)(x - 13) = 0$.
Since integer is positive, $x = 13$.
✅ The integers are 13 and 14.

(iii) Let base = $x$ cm. Altitude = $x - 7$ cm.
$x^2 + (x - 7)^2 = 13^2 \Rightarrow x^2 + x^2 - 14x + 49 = 169$
$2x^2 - 14x - 120 = 0 \Rightarrow x^2 - 7x - 60 = 0$.
Factors of 60 with difference 7 are 12 and 5.
$(x - 12)(x + 5) = 0 \Rightarrow x = 12$ (side cannot be negative).
Altitude = $12 - 7 = 5$ cm.
✅ Base = 12 cm, Altitude = 5 cm.

(iv) Let number of articles = $x$.
Cost per article = $2x + 3$.
Total cost = $x(2x + 3) = 90 \Rightarrow 2x^2 + 3x - 90 = 0$.
Split middle term: $2x^2 + 15x - 12x - 90 = 0$
$x(2x + 15) - 6(2x + 15) = 0 \Rightarrow (2x + 15)(x - 6) = 0$.
$x = 6$ (articles cannot be negative).
Cost = $2(6) + 3 = 15$.
✅ Number of articles = 6, Cost per article = ₹ 15.

Quadratic Equations – Exercise 4.2

Overview

Question 1

Word Problems