Table of Contents
Exercise 5.1 Solutions – Class X Mathematics
These solutions are based on the Telangana State Class X Mathematics textbook, focusing on quadratic equations. Mathematical expressions are rendered using MathJax.
1. Check whether the following are quadratic equations:
(i) \( (x + 1)^2 = 2(x – 3) \)
Expand the equation: \( (x + 1)^2 = x^2 + 2x + 1 \).
So, \( x^2 + 2x + 1 = 2(x – 3) = 2x – 6 \).
Simplify: \( x^2 + 2x + 1 – 2x + 6 = 0 \implies x^2 + 7 = 0 \).
This is a quadratic equation because it has the form \( ax^2 + bx + c = 0 \) with \( a = 1 \), \( b = 0 \), \( c = 7 \).
Conclusion: Yes, it is a quadratic equation.
(ii) \( x^2 – 2x = (-2)(3 – x) \)
Expand the right side: \( (-2)(3 – x) = -6 + 2x \).
So, \( x^2 – 2x = -6 + 2x \).
Simplify: \( x^2 – 2x – 2x + 6 = 0 \implies x^2 – 4x + 6 = 0 \).
This is a quadratic equation (\( a = 1 \), \( b = -4 \), \( c = 6 \)).
Conclusion: Yes, it is a quadratic equation.
(iii) \( (x – 2)(x + 1) = (x – 1)(x + 3) \)
Expand both sides: Left: \( (x – 2)(x + 1) = x^2 + x – 2x – 2 = x^2 – x – 2 \).
Right: \( (x – 1)(x + 3) = x^2 + 3x – x – 3 = x^2 + 2x – 3 \).
So, \( x^2 – x – 2 = x^2 + 2x – 3 \).
Simplify: \( x^2 – x – 2 – x^2 – 2x + 3 = 0 \implies -3x + 1 = 0 \).
This is a linear equation, not a quadratic equation (no \( x^2 \) term).
Conclusion: No, it is not a quadratic equation.
(iv) \( (x – 3)(2x + 1) = x(x + 5) \)
Expand: Left: \( (x – 3)(2x + 1) = 2x^2 + x – 6x – 3 = 2x^2 – 5x – 3 \).
Right: \( x(x + 5) = x^2 + 5x \).
So, \( 2x^2 – 5x – 3 = x^2 + 5x \).
Simplify: \( 2x^2 – 5x – 3 – x^2 – 5x = 0 \implies x^2 – 10x – 3 = 0 \).
This is a quadratic equation (\( a = 1 \), \( b = -10 \), \( c = -3 \)).
Conclusion: Yes, it is a quadratic equation.
(v) \( (2x – 1)(x – 3) = (x + 5)(x – 1) \)
Expand: Left: \( (2x – 1)(x – 3) = 2x^2 – 6x – x + 3 = 2x^2 – 7x + 3 \).
Right: \( (x + 5)(x – 1) = x^2 – x + 5x – 5 = x^2 + 4x – 5 \).
So, \( 2x^2 – 7x + 3 = x^2 + 4x – 5 \).
Simplify: \( 2x^2 – 7x + 3 – x^2 – 4x + 5 = 0 \implies x^2 – 11x + 8 = 0 \).
This is a quadratic equation (\( a = 1 \), \( b = -11 \), \( c = 8 \)).
Conclusion: Yes, it is a quadratic equation.
(vi) \( x^2 + 3x + 1 = (x – 2)^2 \)
Expand: Right: \( (x – 2)^2 = x^2 – 4x + 4 \).
So, \( x^2 + 3x + 1 = x^2 – 4x + 4 \).
Simplify: \( x^2 + 3x + 1 – x^2 + 4x – 4 = 0 \implies 7x – 3 = 0 \).
This is a linear equation, not a quadratic equation.
Conclusion: No, it is not a quadratic equation.
(vii) \( (x + 2)^3 = 2x(x^2 – 1) \)
Expand: Left: \( (x + 2)^3 = x^3 + 3x^2(2) + 3x(2^2) + 2^3 = x^3 + 6x^2 + 12x + 8 \).
Right: \( 2x(x^2 – 1) = 2x^3 – 2x \).
So, \( x^3 + 6x^2 + 12x + 8 = 2x^3 – 2x \).
Simplify: \( x^3 + 6x^2 + 12x + 8 – 2x^3 + 2x = 0 \implies -x^3 + 6x^2 + 14x + 8 = 0 \).
This is a cubic equation (highest degree 3), not a quadratic equation.
Conclusion: No, it is not a quadratic equation.
(viii) \( x^3 – 4x^2 – x + 1 = (x – 2)^3 \)
Expand: Right: \( (x – 2)^3 = x^3 – 3x^2(2) + 3x(2^2) – 2^3 = x^3 – 6x^2 + 12x – 8 \).
So, \( x^3 – 4x^2 – x + 1 = x^3 – 6x^2 + 12x – 8 \).
Simplify: \( x^3 – 4x^2 – x + 1 – x^3 + 6x^2 – 12x + 8 = 0 \implies 2x^2 – 13x + 9 = 0 \).
This is a quadratic equation (\( a = 2 \), \( b = -13 \), \( c = 9 \)).
Conclusion: Yes, it is a quadratic equation.
2. Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length of the plot is one meter more than twice its breadth. We need to find the length and breadth of the plot.
Let the breadth of the plot be \( x \) meters.
Length is one meter more than twice the breadth: \( 2x + 1 \).
Area of the rectangle: \( \text{length} \times \text{breadth} = 528 \).
So, \( x (2x + 1) = 528 \).
Expand: \( 2x^2 + x = 528 \implies 2x^2 + x – 528 = 0 \).
Quadratic equation: \( 2x^2 + x – 528 = 0 \)
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Let the first integer be \( x \).
The next consecutive integer is \( x + 1 \).
Their product is 306: \( x (x + 1) = 306 \).
Expand: \( x^2 + x = 306 \implies x^2 + x – 306 = 0 \).
Quadratic equation: \( x^2 + x – 306 = 0 \)
(iii) Rohan’s mother is 26 years older than him. The product of their ages after 3 years will be 360 years. We need to find Rohan’s present age.
Let Rohan’s present age be \( x \) years.
His mother’s present age is \( x + 26 \).
After 3 years: Rohan’s age = \( x + 3 \), mother’s age = \( x + 29 \).
Product of their ages: \( (x + 3)(x + 29) = 360 \).
Expand: \( x^2 + 29x + 3x + 87 = 360 \implies x^2 + 32x + 87 – 360 = 0 \implies x^2 + 32x – 273 = 0 \).
Quadratic equation: \( x^2 + 32x – 273 = 0 \)
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Let the speed of the train be \( x \) km/h.
Time to travel 480 km at speed \( x \): \( \frac{480}{x} \) hours.
If speed is \( x – 8 \), time taken: \( \frac{480}{x-8} \).
The slower speed takes 3 hours more: \( \frac{480}{x-8} = \frac{480}{x} + 3 \).
Simplify: \( \frac{480}{x-8} – \frac{480}{x} = 3 \implies 480 \left( \frac{x – (x-8)}{x(x-8)} \right) = 3 \implies \frac{480 \cdot 8}{x(x-8)} = 3 \implies \frac{3840}{x^2 – 8x} = 3 \).
\( 3840 = 3(x^2 – 8x) \implies 1280 = x^2 – 8x \implies x^2 – 8x – 1280 = 0 \).
Quadratic equation: \( x^2 – 8x – 1280 = 0 \)
