Table of Contents
Exercise 5.2 Solutions – Class X Mathematics
These solutions are based on the Telangana State Class X Mathematics textbook, focusing on solving quadratic equations by factorisation and applying them to word problems. Mathematical expressions are rendered using MathJax.
1. Find the roots of the following quadratic equations by factorisation:
(i) \( x^2 – 3x – 10 = 0 \)
We need two numbers whose product is \( -10 \cdot 1 = -10 \) and sum is \( -3 \).
Numbers are \( -5 \) and \( 2 \): \( -5 + 2 = -3 \), \( -5 \cdot 2 = -10 \).
Rewrite: \( x^2 – 5x + 2x – 10 = 0 \).
Factor: \( x(x – 5) + 2(x – 5) = 0 \implies (x – 5)(x + 2) = 0 \).
Solve: \( x – 5 = 0 \implies x = 5 \), \( x + 2 = 0 \implies x = -2 \).
Check: \( x = 5 \): \( 5^2 – 3(5) – 10 = 25 – 15 – 10 = 0 \), \( x = -2 \): \( (-2)^2 – 3(-2) – 10 = 4 + 6 – 10 = 0 \).
Roots: \( x = 5, -2 \)
(ii) \( 2x^2 + x – 6 = 0 \)
Product: \( 2 \cdot (-6) = -12 \), sum: \( 1 \).
Numbers: \( 4 \) and \( -3 \): \( 4 + (-3) = 1 \), \( 4 \cdot (-3) = -12 \).
Rewrite: \( 2x^2 + 4x – 3x – 6 = 0 \).
Factor: \( 2x(x + 2) – 3(x + 2) = 0 \implies (x + 2)(2x – 3) = 0 \).
Solve: \( x + 2 = 0 \implies x = -2 \), \( 2x – 3 = 0 \implies x = \frac{3}{2} \).
Check: \( x = -2 \): \( 2(-2)^2 + (-2) – 6 = 8 – 2 – 6 = 0 \), \( x = \frac{3}{2} \): \( 2\left(\frac{3}{2}\right)^2 + \frac{3}{2} – 6 = 2 \cdot \frac{9}{4} + \frac{3}{2} – 6 = \frac{9}{2} + \frac{3}{2} – 6 = 0 \).
Roots: \( x = -2, \frac{3}{2} \)
(iii) \( \sqrt{2} x^2 + 7x + 5\sqrt{2} = 0 \)
Product: \( \sqrt{2} \cdot 5\sqrt{2} = 10 \), sum: \( 7 \).
Numbers: \( 5 \) and \( 2\sqrt{2} \): \( 5 + 2\sqrt{2} = 7 \), \( 5 \cdot 2\sqrt{2} = 10 \).
Rewrite: \( \sqrt{2} x^2 + 5x + 2\sqrt{2} x + 5\sqrt{2} = 0 \).
Factor: \( x(\sqrt{2} x + 5) + \sqrt{2} (\sqrt{2} x + 5) = 0 \implies (\sqrt{2} x + 5)(x + \sqrt{2}) = 0 \).
Solve: \( \sqrt{2} x + 5 = 0 \implies x = -\frac{5}{\sqrt{2}} = -\frac{5\sqrt{2}}{2} \), \( x + \sqrt{2} = 0 \implies x = -\sqrt{2} \).
Check: \( x = -\frac{5\sqrt{2}}{2} \): \( \sqrt{2} \left(-\frac{5\sqrt{2}}{2}\right)^2 + 7 \left(-\frac{5\sqrt{2}}{2}\right) + 5\sqrt{2} = \sqrt{2} \cdot \frac{25 \cdot 2}{4} – \frac{35\sqrt{2}}{2} + 5\sqrt{2} = \frac{25}{2} – \frac{30\sqrt{2}}{2} = 0 \).
Roots: \( x = -\sqrt{2}, -\frac{5\sqrt{2}}{2} \)
(iv) \( 2x^2 – x + \frac{1}{8} = 0 \)
Multiply through by 8: \( 16x^2 – 8x + 1 = 0 \).
Product: \( 16 \cdot 1 = 16 \), sum: \( -8 \).
Numbers: \( -4 \) and \( -4 \): \( -4 + (-4) = -8 \), \( -4 \cdot (-4) = 16 \).
Rewrite: \( 16x^2 – 4x – 4x + 1 = 0 \).
Factor: \( 4x(4x – 1) – 1(4x – 1) = 0 \implies (4x – 1)^2 = 0 \).
Solve: \( 4x – 1 = 0 \implies x = \frac{1}{4} \).
Check: \( 2\left(\frac{1}{4}\right)^2 – \frac{1}{4} + \frac{1}{8} = 2 \cdot \frac{1}{16} – \frac{1}{4} + \frac{1}{8} = \frac{1}{8} – \frac{1}{4} + \frac{1}{8} = 0 \).
Roots: \( x = \frac{1}{4} \) (repeated root)
(v) \( 100x^2 – 20x + 1 = 0 \)
Product: \( 100 \cdot 1 = 100 \), sum: \( -20 \).
Numbers: \( -10 \) and \( -10 \): \( -10 + (-10) = -20 \), \( -10 \cdot (-10) = 100 \).
Rewrite: \( 100x^2 – 10x – 10x + 1 = 0 \).
Factor: \( 10x(10x – 1) – 1(10x – 1) = 0 \implies (10x – 1)^2 = 0 \).
Solve: \( 10x – 1 = 0 \implies x = \frac{1}{10} \).
Check: \( 100\left(\frac{1}{10}\right)^2 – 20\left(\frac{1}{10}\right) + 1 = 100 \cdot \frac{1}{100} – 2 + 1 = 1 – 2 + 1 = 0 \).
Roots: \( x = \frac{1}{10} \) (repeated root)
(vi) \( x(x + 4) = 12 \)
Expand: \( x^2 + 4x = 12 \implies x^2 + 4x – 12 = 0 \).
Product: \( 1 \cdot (-12) = -12 \), sum: \( 4 \).
Numbers: \( 6 \) and \( -2 \): \( 6 + (-2) = 4 \), \( 6 \cdot (-2) = -12 \).
Rewrite: \( x^2 + 6x – 2x – 12 = 0 \).
Factor: \( x(x + 6) – 2(x + 6) = 0 \implies (x + 6)(x – 2) = 0 \).
Solve: \( x + 6 = 0 \implies x = -6 \), \( x – 2 = 0 \implies x = 2 \).
Check: \( x = 2 \): \( 2(2 + 4) = 12 \), \( x = -6 \): \( -6(-6 + 4) = -6(-2) = 12 \).
Roots: \( x = 2, -6 \)
(vii) \( 3x^2 – 5x + 2 = 0 \)
Product: \( 3 \cdot 2 = 6 \), sum: \( -5 \).
Numbers: \( -3 \) and \( -2 \): \( -3 + (-2) = -5 \), \( -3 \cdot (-2) = 6 \).
Rewrite: \( 3x^2 – 3x – 2x + 2 = 0 \).
Factor: \( 3x(x – 1) – 2(x – 1) = 0 \implies (x – 1)(3x – 2) = 0 \).
Solve: \( x – 1 = 0 \implies x = 1 \), \( 3x – 2 = 0 \implies x = \frac{2}{3} \).
Check: \( x = 1 \): \( 3(1)^2 – 5(1) + 2 = 3 – 5 + 2 = 0 \), \( x = \frac{2}{3} \): \( 3\left(\frac{2}{3}\right)^2 – 5\left(\frac{2}{3}\right) + 2 = 3 \cdot \frac{4}{9} – \frac{10}{3} + 2 = \frac{4}{3} – \frac{10}{3} + 2 = 0 \).
Roots: \( x = 1, \frac{2}{3} \)
(viii) \( x – \frac{3}{x} = 2 \) (\( x \neq 0 \))
Multiply through by \( x \): \( x^2 – 3 = 2x \implies x^2 – 2x – 3 = 0 \).
Product: \( 1 \cdot (-3) = -3 \), sum: \( -2 \).
Numbers: \( -3 \) and \( 1 \): \( -3 + 1 = -2 \), \( -3 \cdot 1 = -3 \).
Rewrite: \( x^2 – 3x + x – 3 = 0 \).
Factor: \( x(x – 3) + 1(x – 3) = 0 \implies (x – 3)(x + 1) = 0 \).
Solve: \( x – 3 = 0 \implies x = 3 \), \( x + 1 = 0 \implies x = -1 \).
Check: \( x = 3 \): \( 3 – \frac{3}{3} = 2 \), \( x = -1 \): \( -1 – \frac{3}{-1} = -1 + 3 = 2 \).
Roots: \( x = 3, -1 \)
(ix) \( 3(x – 4)^2 – 5(x – 4) = 12 \)
Substitute \( u = x – 4 \). Then: \( 3u^2 – 5u = 12 \implies 3u^2 – 5u – 12 = 0 \).
Product: \( 3 \cdot (-12) = -36 \), sum: \( -5 \).
Numbers: \( -9 \) and \( 4 \): \( -9 + 4 = -5 \), \( -9 \cdot 4 = -36 \).
Rewrite: \( 3u^2 – 9u + 4u – 12 = 0 \).
Factor: \( 3u(u – 3) + 4(u – 3) = 0 \implies (u – 3)(3u + 4) = 0 \).
Solve: \( u – 3 = 0 \implies u = 3 \), \( 3u + 4 = 0 \implies u = -\frac{4}{3} \).
Back-substitute: \( x – 4 = 3 \implies x = 7 \), \( x – 4 = -\frac{4}{3} \implies x = -\frac{4}{3} + 4 = \frac{8}{3} \).
Check: \( x = 7 \): \( 3(7 – 4)^2 – 5(7 – 4) = 3(3)^2 – 5(3) = 27 – 15 = 12 \).
Roots: \( x = 7, \frac{8}{3} \)
2. Find two numbers whose sum is 27 and product is 182.
Let the numbers be \( x \) and \( 27 – x \).
Product: \( x(27 – x) = 182 \implies 27x – x^2 = 182 \implies x^2 – 27x + 182 = 0 \).
Product: \( 1 \cdot 182 = 182 \), sum: \( -27 \).
Numbers: \( -14 \) and \( -13 \): \( -14 + (-13) = -27 \), \( -14 \cdot (-13) = 182 \).
Rewrite: \( x^2 – 14x – 13x + 182 = 0 \).
Factor: \( x(x – 14) – 13(x – 14) = 0 \implies (x – 14)(x – 13) = 0 \).
Solve: \( x = 14 \), \( x = 13 \).
Numbers: \( 14 \) and \( 27 – 14 = 13 \).
Check: Sum: \( 14 + 13 = 27 \), Product: \( 14 \cdot 13 = 182 \).
Numbers: 14 and 13
3. Find two consecutive positive integers, sum of whose squares is 613.
Let the integers be \( x \) and \( x + 1 \).
Sum of squares: \( x^2 + (x + 1)^2 = 613 \).
Expand: \( x^2 + x^2 + 2x + 1 = 613 \implies 2x^2 + 2x + 1 – 613 = 0 \implies 2x^2 + 2x – 612 = 0 \implies x^2 + x – 306 = 0 \).
Product: \( -306 \), sum: \( 1 \).
Numbers: \( 18 \) and \( -17 \): \( 18 + (-17) = 1 \), \( 18 \cdot (-17) = -306 \).
Rewrite: \( x^2 + 18x – 17x – 306 = 0 \).
Factor: \( x(x + 18) – 17(x + 18) = 0 \implies (x + 18)(x – 17) = 0 \).
Solve: \( x = -18 \), \( x = 17 \). Since positive integers are required, take \( x = 17 \).
Integers: \( 17 \) and \( 18 \).
Check: \( 17^2 + 18^2 = 289 + 324 = 613 \).
Integers: 17 and 18
4. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Let the base be \( x \) cm, altitude = \( x – 7 \).
By Pythagoras: \( x^2 + (x – 7)^2 = 13^2 \).
Expand: \( x^2 + x^2 – 14x + 49 = 169 \implies 2x^2 – 14x + 49 – 169 = 0 \implies 2x^2 – 14x – 120 = 0 \implies x^2 – 7x – 60 = 0 \).
Product: \( -60 \), sum: \( -7 \).
Numbers: \( -12 \) and \( 5 \): \( -12 + 5 = -7 \), \( -12 \cdot 5 = -60 \).
Rewrite: \( x^2 – 12x + 5x – 60 = 0 \).
Factor: \( x(x – 12) + 5(x – 12) = 0 \implies (x – 12)(x + 5) = 0 \).
Solve: \( x = 12 \), \( x = -5 \). Since length is positive, \( x = 12 \).
Base: \( 12 \) cm, altitude: \( 12 – 7 = 5 \) cm.
Check: \( 12^2 + 5^2 = 144 + 25 = 169 = 13^2 \).
Sides: Base = 12 cm, Altitude = 5 cm
5. 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.
Let the number of articles be \( x \).
Cost of each article: \( 2x + 3 \).
Total cost: \( x (2x + 3) = 90 \implies 2x^2 + 3x – 90 = 0 \).
Product: \( 2 \cdot (-90) = -180 \), sum: \( 3 \).
Numbers: \( 15 \) and \( -12 \): \( 15 + (-12) = 3 \), \( 15 \cdot (-12) = -180 \).
Rewrite: \( 2x^2 + 15x – 12x – 90 = 0 \).
Factor: \( x(2x + 15) – 6(2x + 15) = 0 \implies (2x + 15)(x – 6) = 0 \).
Solve: \( 2x + 15 = 0 \implies x = -\frac{15}{2} \), \( x – 6 = 0 \implies x = 6 \). Take \( x = 6 \).
Number of articles: \( 6 \), cost per article: \( 2(6) + 3 = 15 \).
Check: Total cost: \( 6 \cdot 15 = 90 \).
Articles: 6, Cost per article: Rs 15
6. Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40 square meters.
Let length be \( x \) m, breadth be \( y \) m.
Perimeter: \( 2(x + y) = 28 \implies x + y = 14 \implies y = 14 – x \).
Area: \( x y = 40 \).
Substitute: \( x (14 – x) = 40 \implies 14x – x^2 = 40 \implies x^2 – 14x + 40 = 0 \).
Product: \( 40 \), sum: \( -14 \).
Numbers: \( -10 \) and \( -4 \): \( -10 + (-4) = -14 \), \( -10 \cdot (-4) = 40 \).
Rewrite: \( x^2 – 10x – 4x + 40 = 0 \).
Factor: \( x(x – 10) – 4(x – 10) = 0 \implies (x – 10)(x – 4) = 0 \).
Solve: \( x = 10 \), \( x = 4 \).
If \( x = 10 \), \( y = 14 – 10 = 4 \). If \( x = 4 \), \( y = 10 \).
Check: Perimeter: \( 2(10 + 4) = 28 \), Area: \( 10 \cdot 4 = 40 \).
Dimensions: 10 m and 4 m
7. The base of a triangle is 4 cm longer than its altitude. If the area of the triangle is 48 sq.cm then find its base and altitude.
Let the altitude be \( x \) cm, base = \( x + 4 \).
Area: \( \frac{1}{2} \times \text{base} \times \text{altitude} = 48 \implies \frac{1}{2} (x + 4) x = 48 \implies x(x + 4) = 96 \).
\( x^2 + 4x – 96 = 0 \).
Product: \( -96 \), sum: \( 4 \).
Numbers: \( 12 \) and \( -8 \): \( 12 + (-8) = 4 \), \( 12 \cdot (-8) = -96 \).
Rewrite: \( x^2 + 12x – 8x – 96 = 0 \).
Factor: \( x(x + 12) – 8(x + 12) = 0 \implies (x + 12)(x – 8) = 0 \).
Solve: \( x = -12 \), \( x = 8 \). Take \( x = 8 \).
Altitude: \( 8 \) cm, base: \( 8 + 4 = 12 \) cm.
Check: Area: \( \frac{1}{2} \times 12 \times 8 = 48 \).
Base: 12 cm, Altitude: 8 cm
8. Two trains leave a railway station at the same time. The first train travels towards west and the second train towards north. The first train travels 5 km/hr faster than the second train. If after two hours they are 50 km apart, find the average speed of each train.
Let the speed of the second train (north) be \( x \) km/h, first train (west) = \( x + 5 \) km/h.
After 2 hours: Distance west = \( 2(x + 5) \), distance north = \( 2x \).
Distance apart (right triangle): \( \sqrt{(2x)^2 + (2(x + 5))^2} = 50 \).
Simplify: \( 4x^2 + 4(x^2 + 10x + 25) = 2500 \implies 4x^2 + 4x^2 + 40x + 100 = 2500 \implies 8x^2 + 40x – 2400 = 0 \implies x^2 + 5x – 300 = 0 \).
Product: \( -300 \), sum: \( 5 \).
Numbers: \( 20 \) and \( -15 \): \( 20 + (-15) = 5 \), \( 20 \cdot (-15) = -300 \).
Rewrite: \( x^2 + 20x – 15x – 300 = 0 \).
Factor: \( x(x + 20) – 15(x + 20) = 0 \implies (x + 20)(x – 15) = 0 \).
Solve: \( x = -20 \), \( x = 15 \). Take \( x = 15 \).
Second train: \( 15 \) km/h, first train: \( 15 + 5 = 20 \) km/h.
Check: Distances: \( 2 \cdot 15 = 30 \), \( 2 \cdot 20 = 40 \). \( \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = 50 \).
Speeds: First train (west): 20 km/h, Second train (north): 15 km/h
9. In a class of 60 students, each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys. If the total money then collected was Rs 1600, how many boys were there in the class?
Let the number of boys be \( x \), girls = \( 60 – x \).
Boys contribute: \( x \cdot (60 – x) \), girls contribute: \( (60 – x) \cdot x \).
Total: \( x(60 – x) + (60 – x)x = 2x(60 – x) = 1600 \).
Simplify: \( 2x(60 – x) = 1600 \implies 120x – 2x^2 = 1600 \implies x^2 – 60x + 800 = 0 \).
Product: \( 800 \), sum: \( -60 \).
Numbers: \( -40 \) and \( -20 \): \( -40 + (-20) = -60 \), \( -40 \cdot (-20) = 800 \).
Rewrite: \( x^2 – 40x – 20x + 800 = 0 \).
Factor: \( x(x – 40) – 20(x – 40) = 0 \implies (x – 40)(x – 20) = 0 \).
Solve: \( x = 40 \), \( x = 20 \).
If \( x = 40 \), girls = \( 20 \). If \( x = 20 \), girls = \( 40 \). Both are symmetric.
Check: \( 40 \cdot 20 + 20 \cdot 40 = 800 + 800 = 1600 \).
Number of boys: 40 or 20 (both possible, but typically \( 40 \) is chosen as “boys”)
10. A motor boat heads upstream a distance of 24 km in a river whose current is running at 3 km per hour. The trip up and back takes 6 hours. Assuming that the motor boat maintained a constant speed, what was its speed in still water?
Let the boat’s speed in still water be \( x \) km/h.
Upstream speed: \( x – 3 \), downstream speed: \( x + 3 \).
Time upstream: \( \frac{24}{x-3} \), time downstream: \( \frac{24}{x+3} \).
Total time: \( \frac{24}{x-3} + \frac{24}{x+3} = 6 \).
Simplify: \( 24 \left( \frac{(x+3) + (x-3)}{(x-3)(x+3)} \right) = 6 \implies 24 \cdot \frac{2x}{x^2 – 9} = 6 \implies 48x = 6(x^2 – 9) \implies 8x = x^2 – 9 \implies x^2 – 8x – 9 = 0 \).
Product: \( -9 \), sum: \( -8 \).
Numbers: \( -9 \) and \( 1 \): \( -9 + 1 = -8 \), \( -9 \cdot 1 = -9 \).
Rewrite: \( x^2 – 9x + x – 9 = 0 \).
Factor: \( x(x – 9) + 1(x – 9) = 0 \implies (x – 9)(x + 1) = 0 \).
Solve: \( x = 9 \), \( x = -1 \). Take \( x = 9 \).
Check: \( \frac{24}{9-3} + \frac{24}{9+3} = \frac{24}{6} + \frac{24}{12} = 4 + 2 = 6 \).
Speed in still water: 9 km/h
