Class 10 Maths Part 1 Chapter 2 Practice Set 2.3 - Quadratic Equations Solutions (Maharashtra Board)
Welcome to the comprehensive Textbook Exercise Solutions for Maharashtra Board Class 10 Maths Part 1. In this post, we will cover Chapter 2: Quadratic Equations, specifically focusing on Practice Set 2.3. This exercise is essential as it deals with solving quadratic equations using the completing the square method. Our step-by-step guide will help you understand the concepts clearly and score full marks in your board exams!
Exam Tip: When using the "Completing the Square" method, always ensure the coefficient of \( x^2 \) (or the leading variable) is exactly 1 before proceeding. The formula for the third term (k) is \( \left( \frac{1}{2} \times \text{coefficient of } x \right)^2 \). Add and subtract this term to form a perfect square binomial.
Question 1: Solve the following quadratic equations by completing the square method.
1. \( x^{2} + x - 20 = 0 \)
Solution & Calculation:
\( x^{2} + x - 20 = 0 \)
Let \( x^{2} + x + k = (x + a)^{2} \)
\( x^{2} + x + k = x^{2} + 2ax + a^{2} \)
Comparing the coefficients, we get:
\( 1 = 2a \) and \( k = a^{2} \)
\( a = \frac{1}{2} \) and \( k = \left(\frac{1}{2}\right)^{2} = \frac{1}{4} \)
Now, substitute \( k \) back into the original equation context:
\( x^{2} + x + \frac{1}{4} - \frac{1}{4} - 20 = 0 \)
\( \left(x + \frac{1}{2}\right)^{2} - \left(\frac{1 + 80}{4}\right) = 0 \)
\( \left(x + \frac{1}{2}\right)^{2} - \frac{81}{4} = 0 \)
\( \left(x + \frac{1}{2}\right)^{2} = \frac{81}{4} \)
Taking the square root of both sides, we get:
\( x + \frac{1}{2} = \pm \frac{9}{2} \)
\( x + \frac{1}{2} = \frac{9}{2} \) or \( x + \frac{1}{2} = -\frac{9}{2} \)
\( x = \frac{9}{2} - \frac{1}{2} \) or \( x = -\frac{9}{2} - \frac{1}{2} \)
\( x = \frac{8}{2} = 4 \) or \( x = -\frac{10}{2} = -5 \)
Final Answer: The roots of the given quadratic equation are 4 and -5.
2. \( x^{2} + 2x - 5 = 0 \)
Solution & Calculation:
\( x^{2} + 2x - 5 = 0 \)
Let \( x^{2} + 2x + k = (x + a)^{2} \)
\( x^{2} + 2x + k = x^{2} + 2ax + a^{2} \)
Comparing the coefficients, we get:
\( 2 = 2a \) and \( k = a^{2} \)
\( a = 1 \) and \( k = (1)^{2} = 1 \)
Now, substitute \( k \):
\( x^{2} + 2x + 1 - 1 - 5 = 0 \)
\( (x + 1)^{2} - 6 = 0 \)
\( (x + 1)^{2} = 6 \)
Taking the square root of both sides, we get:
\( x + 1 = \pm \sqrt{6} \)
\( x + 1 = \sqrt{6} \) or \( x + 1 = -\sqrt{6} \)
\( x = \sqrt{6} - 1 \) or \( x = -\sqrt{6} - 1 \)
Final Answer: The roots of the given quadratic equation are \( \sqrt{6} - 1 \) and \( -\sqrt{6} - 1 \).
3. \( m^{2} - 5m = -3 \)
Solution & Calculation:
\( m^{2} - 5m = -3 \)
\( m^{2} - 5m + 3 = 0 \)
Let \( m^{2} - 5m + k = (m + a)^{2} \)
\( m^{2} - 5m + k = m^{2} + 2am + a^{2} \)
Comparing the coefficients, we get:
\( -5 = 2a \) and \( k = a^{2} \)
\( a = \frac{-5}{2} \) and \( k = \left(\frac{-5}{2}\right)^{2} = \frac{25}{4} \)
Now, substitute \( k \):
\( m^{2} - 5m + \frac{25}{4} - \frac{25}{4} + 3 = 0 \)
\( \left(m - \frac{5}{2}\right)^{2} + \left(\frac{-25 + 12}{4}\right) = 0 \)
\( \left(m - \frac{5}{2}\right)^{2} - \frac{13}{4} = 0 \)
\( \left(m - \frac{5}{2}\right)^{2} = \frac{13}{4} \)
Taking the square root of both sides, we get:
\( m - \frac{5}{2} = \pm \frac{\sqrt{13}}{2} \)
\( m - \frac{5}{2} = \frac{\sqrt{13}}{2} \) or \( m - \frac{5}{2} = -\frac{\sqrt{13}}{2} \)
\( m = \frac{\sqrt{13} + 5}{2} \) or \( m = \frac{-\sqrt{13} + 5}{2} \)
Final Answer: The roots of the given quadratic equation are \( \frac{\sqrt{13} + 5}{2} \) and \( \frac{-\sqrt{13} + 5}{2} \).
4. \( 9y^{2} - 12y + 2 = 0 \)
Solution & Calculation:
\( y^{2} - \frac{4}{3}y + \frac{2}{9} = 0 \) (Dividing both sides by 9)
Let \( y^{2} - \frac{4}{3}y + k = (y + a)^{2} \)
\( y^{2} - \frac{4}{3}y + k = y^{2} + 2ay + a^{2} \)
Comparing the coefficients, we get:
\( -\frac{4}{3} = 2a \) and \( k = a^{2} \)
\( a = -\frac{2}{3} \) and \( k = \left(-\frac{2}{3}\right)^{2} = \frac{4}{9} \)
Now, substitute \( k \):
\( y^{2} - \frac{4}{3}y + \frac{4}{9} - \frac{4}{9} + \frac{2}{9} = 0 \)
\( \left(y - \frac{2}{3}\right)^{2} - \frac{2}{9} = 0 \)
\( \left(y - \frac{2}{3}\right)^{2} = \frac{2}{9} \)
Taking the square root of both sides, we get:
\( y - \frac{2}{3} = \pm \frac{\sqrt{2}}{3} \)
\( y - \frac{2}{3} = \frac{\sqrt{2}}{3} \) or \( y - \frac{2}{3} = -\frac{\sqrt{2}}{3} \)
\( y = \frac{\sqrt{2} + 2}{3} \) or \( y = \frac{-\sqrt{2} + 2}{3} \)
Final Answer: The roots of the given quadratic equation are \( \frac{\sqrt{2} + 2}{3} \) and \( \frac{-\sqrt{2} + 2}{3} \).
5. \( 2y^{2} + 9y + 10 = 0 \)
Solution & Calculation:
\( y^{2} + \frac{9}{2}y + 5 = 0 \) (Dividing both sides by 2)
Let \( y^{2} + \frac{9}{2}y + k = (y + a)^{2} \)
\( y^{2} + \frac{9}{2}y + k = y^{2} + 2ay + a^{2} \)
Comparing the coefficients, we get:
\( \frac{9}{2} = 2a \) and \( k = a^{2} \)
\( a = \frac{9}{4} \) and \( k = \left(\frac{9}{4}\right)^{2} = \frac{81}{16} \)
Now, substitute \( k \):
\( y^{2} + \frac{9}{2}y + \frac{81}{16} - \frac{81}{16} + 5 = 0 \)
\( \left(y + \frac{9}{4}\right)^{2} + \left(\frac{-81 + 80}{16}\right) = 0 \)
\( \left(y + \frac{9}{4}\right)^{2} - \frac{1}{16} = 0 \)
\( \left(y + \frac{9}{4}\right)^{2} = \frac{1}{16} \)
Taking the square root of both sides, we get:
\( y + \frac{9}{4} = \pm \frac{1}{4} \)
\( y + \frac{9}{4} = \frac{1}{4} \) or \( y + \frac{9}{4} = -\frac{1}{4} \)
\( y = \frac{1}{4} - \frac{9}{4} \) or \( y = -\frac{1}{4} - \frac{9}{4} \)
\( y = \frac{-8}{4} = -2 \) or \( y = -\frac{10}{4} = -\frac{5}{2} \)
Final Answer: The roots of the given quadratic equation are -2 and \( -\frac{5}{2} \).
6. \( 5x^{2} = 4x + 7 \)
Solution & Calculation:
\( 5x^{2} - 4x - 7 = 0 \)
\( x^{2} - \frac{4}{5}x - \frac{7}{5} = 0 \) (Dividing both sides by 5)
Let \( x^{2} - \frac{4}{5}x + k = (x + a)^{2} \)
\( x^{2} - \frac{4}{5}x + k = x^{2} + 2ax + a^{2} \)
Comparing the coefficients, we get:
\( -\frac{4}{5} = 2a \) and \( k = a^{2} \)
\( a = -\frac{2}{5} \) and \( k = \left(-\frac{2}{5}\right)^{2} = \frac{4}{25} \)
Now, substitute \( k \):
\( x^{2} - \frac{4}{5}x + \frac{4}{25} - \frac{4}{25} - \frac{7}{5} = 0 \)
\( \left(x - \frac{2}{5}\right)^{2} - \left(\frac{4 + 35}{25}\right) = 0 \)
\( \left(x - \frac{2}{5}\right)^{2} - \frac{39}{25} = 0 \)
\( \left(x - \frac{2}{5}\right)^{2} = \frac{39}{25} \)
Taking the square root of both sides, we get:
\( x - \frac{2}{5} = \pm \frac{\sqrt{39}}{5} \)
\( x - \frac{2}{5} = \frac{\sqrt{39}}{5} \) or \( x - \frac{2}{5} = -\frac{\sqrt{39}}{5} \)
\( x = \frac{2 + \sqrt{39}}{5} \) or \( x = \frac{2 - \sqrt{39}}{5} \)
Final Answer: The roots of the given quadratic equation are \( \frac{2 + \sqrt{39}}{5} \) and \( \frac{2 - \sqrt{39}}{5} \).
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