OMTEX CLASSES: Practice Set 2.2

Welcome to the complete, step-by-step Textbook Exercise Solutions for Maharashtra Board Class 10 Maths Part 1. In this post, we will cover Chapter 2: Quadratic Equations, focusing entirely on Practice Set 2.2. These Chapter Notes and solutions are designed to help you master solving quadratic equations by factorization, ensuring you score full marks in your board exams!

Exam Tip: The Zero Product Property
The factorization method relies heavily on this core mathematical rule: If the product of two numbers is zero, then at least one of them must be zero. In exam terms, if \( (a)(b) = 0 \), then either \( a = 0 \) or \( b = 0 \). Keep this principle in mind for every factorization sum!

Practice Set 2.2

Question 1. Solve the following quadratic equations by factorisation.

Solution i.

Given:
\[ x^2 - 15x + 54 = 0 \]
Calculation: \[ x^2 - 9x - 6x + 54 = 0 \] \[ x(x - 9) - 6(x - 9) = 0 \] \[ (x - 9)(x - 6) = 0 \] By using the zero product property: \[ x - 9 = 0 \quad \text{or} \quad x - 6 = 0 \] \[ x = 9 \quad \text{or} \quad x = 6 \]
Final Answer:
The roots of the given quadratic equation are 9 and 6.

Solution ii.

Given:
\[ x^2 + x - 20 = 0 \]
Calculation: \[ x^2 + 5x - 4x - 20 = 0 \] \[ x(x + 5) - 4(x + 5) = 0 \] \[ (x + 5)(x - 4) = 0 \] By using the zero product property: \[ x + 5 = 0 \quad \text{or} \quad x - 4 = 0 \] \[ x = -5 \quad \text{or} \quad x = 4 \]
Final Answer:
The roots of the given quadratic equation are -5 and 4.

Solution iii.

Given:
\[ 2y^2 + 27y + 13 = 0 \]
Calculation: \[ 2y^2 + 26y + y + 13 = 0 \] \[ 2y(y + 13) + 1(y + 13) = 0 \] \[ (y + 13)(2y + 1) = 0 \] By using the zero product property: \[ y + 13 = 0 \quad \text{or} \quad 2y + 1 = 0 \] \[ y = -13 \quad \text{or} \quad 2y = -1 \] \[ y = -13 \quad \text{or} \quad y = -\frac{1}{2} \]
Final Answer:
The roots of the given quadratic equation are -13 and \( -\frac{1}{2} \).

Solution iv.

Given:
\[ 5m^2 = 22m + 15 \]
Calculation:
First, bring all terms to one side to form the standard quadratic equation \( ax^2 + bx + c = 0 \): \[ 5m^2 - 22m - 15 = 0 \] \[ 5m^2 - 25m + 3m - 15 = 0 \] \[ 5m(m - 5) + 3(m - 5) = 0 \] \[ (m - 5)(5m + 3) = 0 \] By using the zero product property: \[ m - 5 = 0 \quad \text{or} \quad 5m + 3 = 0 \] \[ m = 5 \quad \text{or} \quad 5m = -3 \] \[ m = 5 \quad \text{or} \quad m = -\frac{3}{5} \]
Final Answer:
The roots of the given quadratic equation are 5 and \( -\frac{3}{5} \).

Solution v.

Given:
\[ 2x^2 - 2x + \frac{1}{2} = 0 \]
Calculation:
Multiplying both sides by 2 to remove the fraction: \[ 4x^2 - 4x + 1 = 0 \] \[ 4x^2 - 2x - 2x + 1 = 0 \] \[ 2x(2x - 1) - 1(2x - 1) = 0 \] \[ (2x - 1)(2x - 1) = 0 \] By using the zero product property: \[ 2x - 1 = 0 \quad \text{or} \quad 2x - 1 = 0 \] \[ 2x = 1 \quad \text{or} \quad 2x = 1 \] \[ x = \frac{1}{2} \quad \text{or} \quad x = \frac{1}{2} \]
Final Answer:
The roots of the given quadratic equation are \( \frac{1}{2} \) and \( \frac{1}{2} \).

Solution vi.

Given:
\[ 6x - \frac{2}{x} = 1 \]
Calculation:
Multiplying both sides by \( x \) to remove the denominator: \[ 6x^2 - 2 = x \] Rearranging to standard form: \[ 6x^2 - x - 2 = 0 \] \[ 6x^2 - 4x + 3x - 2 = 0 \] \[ 2x(3x - 2) + 1(3x - 2) = 0 \] \[ (3x - 2)(2x + 1) = 0 \] By using the zero product property: \[ 3x - 2 = 0 \quad \text{or} \quad 2x + 1 = 0 \] \[ 3x = 2 \quad \text{or} \quad 2x = -1 \] \[ x = \frac{2}{3} \quad \text{or} \quad x = -\frac{1}{2} \]
Final Answer:
The roots of the given quadratic equation are \( \frac{2}{3} \) and \( -\frac{1}{2} \).

Solution vii.

Given:
\[ \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \]
Calculation: \[ \sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0 \] \[ x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0 \] \[ (\sqrt{2}x + 5)(x + \sqrt{2}) = 0 \] By using the zero product property: \[ \sqrt{2}x + 5 = 0 \quad \text{or} \quad x + \sqrt{2} = 0 \] \[ \sqrt{2}x = -5 \quad \text{or} \quad x = -\sqrt{2} \] \[ x = \frac{-5}{\sqrt{2}} \quad \text{or} \quad x = -\sqrt{2} \]
Final Answer:
The roots of the given quadratic equation are \( \frac{-5}{\sqrt{2}} \) and \( -\sqrt{2} \).

Solution viii.

Given:
\[ 3x^2 - 2\sqrt{6}x + 2 = 0 \]
Calculation: \[ 3x^2 - \sqrt{6}x - \sqrt{6}x + 2 = 0 \] Taking \( \sqrt{3}x \) common from the first two terms, and \( \sqrt{2} \) from the last two: \[ \sqrt{3}x(\sqrt{3}x - \sqrt{2}) - \sqrt{2}(\sqrt{3}x - \sqrt{2}) = 0 \] \[ (\sqrt{3}x - \sqrt{2})(\sqrt{3}x - \sqrt{2}) = 0 \] By using the zero product property: \[ \sqrt{3}x - \sqrt{2} = 0 \quad \text{or} \quad \sqrt{3}x - \sqrt{2} = 0 \] \[ \sqrt{3}x = \sqrt{2} \quad \text{or} \quad \sqrt{3}x = \sqrt{2} \] \[ x = \frac{\sqrt{2}}{\sqrt{3}} \quad \text{or} \quad x = \frac{\sqrt{2}}{\sqrt{3}} \]
Final Answer:
The roots of the given quadratic equation are \( \frac{\sqrt{2}}{\sqrt{3}} \) and \( \frac{\sqrt{2}}{\sqrt{3}} \).

Solution ix.

Given:
\[ 2m(m - 24) = 50 \]
Calculation: \[ 2m^2 - 48m = 50 \] \[ 2m^2 - 48m - 50 = 0 \] Dividing both sides by 2 to simplify: \[ m^2 - 24m - 25 = 0 \] \[ m^2 - 25m + m - 25 = 0 \] \[ m(m - 25) + 1(m - 25) = 0 \] \[ (m - 25)(m + 1) = 0 \] By using the zero product property: \[ m - 25 = 0 \quad \text{or} \quad m + 1 = 0 \] \[ m = 25 \quad \text{or} \quad m = -1 \]
Final Answer:
The roots of the given quadratic equation are 25 and -1.

Solution x.

Given:
\[ 25m^2 = 9 \]
Calculation: \[ 25m^2 - 9 = 0 \] Expressing the equation in the form of \( a^2 - b^2 \): \[ (5m)^2 - (3)^2 = 0 \] Using the identity \( a^2 - b^2 = (a + b)(a - b) \): \[ (5m + 3)(5m - 3) = 0 \] By using the zero product property: \[ 5m + 3 = 0 \quad \text{or} \quad 5m - 3 = 0 \] \[ 5m = -3 \quad \text{or} \quad 5m = 3 \] \[ m = -\frac{3}{5} \quad \text{or} \quad m = \frac{3}{5} \]
Final Answer:
The roots of the given quadratic equation are \( -\frac{3}{5} \) and \( \frac{3}{5} \).

Solution xi.

Given:
\[ 7m^2 = 21m \]
Calculation: \[ 7m^2 - 21m = 0 \] Dividing both sides by 7 to simplify: \[ m^2 - 3m = 0 \] Taking \( m \) common: \[ m(m - 3) = 0 \] By using the zero product property: \[ m = 0 \quad \text{or} \quad m - 3 = 0 \] \[ m = 0 \quad \text{or} \quad m = 3 \]
Final Answer:
The roots of the given quadratic equation are 0 and 3.

Solution xii.

Given:
\[ m^2 - 11 = 0 \]
Calculation: Expressing 11 as a square to apply the difference of squares identity: \[ m^2 - (\sqrt{11})^2 = 0 \] Using the identity \( a^2 - b^2 = (a + b)(a - b) \): \[ (m + \sqrt{11})(m - \sqrt{11}) = 0 \] By using the zero product property: \[ m + \sqrt{11} = 0 \quad \text{or} \quad m - \sqrt{11} = 0 \] \[ m = -\sqrt{11} \quad \text{or} \quad m = \sqrt{11} \]
Final Answer:
The roots of the given quadratic equation are \( -\sqrt{11} \) and \( \sqrt{11} \).

Class 10 Maths Part 1 Practice Set 2.2 Solutions - Maharashtra Board