Maharashtra Board Class 10 Maths Chapter 2 Quadratic Equations Set 2.2 Solutions

Question 1. Solve the following quadratic equations by factorisation.

(i) x² – 15x + 54 = 0
Answer:
\(x^2 - 15x + 54 = 0\)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र मध्य पद को विभाजित करने की प्रक्रिया को दर्शाता है। इसमें 54 को ऐसे दो गुणनखंडों -9 और -6 में विभाजित किया गया है, जिनका गुणनफल 54 है और योगफल -15 है, जो कि समीकरण में x का गुणांक है।
\(x^2 - 9x - 6x + 54 = 0\)
\(x(x - 9) - 6(x - 9) = 0\)
\((x - 9)(x - 6) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(x - 9 = 0\) or \(x - 6 = 0\)
\(x = 9\) or \(x = 6\)
The roots of the given quadratic equation are 9 and 6.
In simple words: To solve this quadratic equation by factorization, we find two numbers whose product is 54 and sum is -15. These numbers are -9 and -6, which allows us to factor the equation into \((x-9)(x-6)=0\), yielding roots \(x=9\) and \(x=6\).

🎯 Exam Tip: Always check your factors by multiplying them to ensure they equal the constant term and add up to the coefficient of the middle term. This step helps in avoiding common calculation errors.

(ii) x² + x − 20 = 0
Answer:
\(x^2 + x - 20 = 0\)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र मध्य पद को विभाजित करने की प्रक्रिया को दर्शाता है। इसमें -20 को ऐसे दो गुणनखंडों 5 और -4 में विभाजित किया गया है, जिनका गुणनफल -20 है और योगफल 1 है, जो कि समीकरण में x का गुणांक है।
\(x^2 + 5x - 4x - 20 = 0\)
\(x(x + 5) - 4(x + 5) = 0\)
\((x + 5)(x - 4) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(x + 5 = 0\) or \(x - 4 = 0\)
\(x = -5\) or \(x = 4\)
The roots of the given quadratic equation are -5 and 4.
In simple words: For the equation \(x^2+x-20=0\), we look for two numbers that multiply to -20 and add to 1. These are 5 and -4, leading to factors \((x+5)(x-4)=0\) and roots \(x=-5\) and \(x=4\).

🎯 Exam Tip: Pay close attention to the signs of the factors when the constant term is negative. One factor must be positive and the other negative to get a negative product.

(iii) 2y² + 27y + 13 = 0
Answer:
\(2y^2 + 27y + 13 = 0\)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र मध्य पद को विभाजित करने की प्रक्रिया को दर्शाता है। इसमें \(2 \times 13 = 26\) को ऐसे दो गुणनखंडों 26 और 1 में विभाजित किया गया है, जिनका गुणनफल 26 है और योगफल 27 है, जो कि समीकरण में y का गुणांक है।
\(2y^2 + 26y + y + 13 = 0\)
\(2y(y + 13) + 1(y + 13) = 0\)
\((y + 13)(2y + 1) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(y + 13 = 0\) or \(2y + 1 = 0\)
\(y = -13\) or \(2y = -1\)
\(y = -13\) or \(y = -\frac{1}{2}\)
The roots of the given quadratic equation are -13 and \(-\frac{1}{2}\).
In simple words: To factor \(2y^2+27y+13=0\), we multiply the first and last coefficients \((2 \times 13 = 26)\) and find factors of 26 that sum to 27 (which are 26 and 1). This splits the middle term, allowing us to factor by grouping to get \((y+13)(2y+1)=0\), leading to roots \(y=-13\) and \(y=-\frac{1}{2}\).

🎯 Exam Tip: When the coefficient of the squared term is not 1, multiply it by the constant term and find factors of this product that sum to the middle term's coefficient. Then factor by grouping.

(iv) 5m² = 22m + 15
Answer:
\(5m^2 = 22m + 15\)
\(5m^2 - 22m - 15 = 0\)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र मध्य पद को विभाजित करने की प्रक्रिया को दर्शाता है। इसमें \(5 \times -15 = -75\) को ऐसे दो गुणनखंडों -25 और +3 में विभाजित किया गया है, जिनका गुणनफल -75 है और योगफल -22 है, जो कि समीकरण में m का गुणांक है।
\(5m^2 - 25m + 3m - 15 = 0\)
\(5m(m - 5) + 3(m - 5) = 0\)
\((m - 5)(5m + 3) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(m - 5 = 0\) or \(5m + 3 = 0\)
\(m = 5\) or \(5m = -3\)
\(m = 5\) or \(m = -\frac{3}{5}\)
The roots of the given quadratic equation are 5 and \(-\frac{3}{5}\).
In simple words: First, rewrite the equation as \(5m^2 - 22m - 15 = 0\). Then, find factors of \((5 \times -15) = -75\) that sum to -22, which are -25 and 3. Factor by grouping to get \((m-5)(5m+3)=0\), yielding roots \(m=5\) and \(m=-\frac{3}{5}\).

🎯 Exam Tip: Always rearrange the quadratic equation into the standard form \(ax^2 + bx + c = 0\) before attempting factorization to ensure correct identification of coefficients.

(v) \(2x^2 - 2x + \frac{1}{2} = 0\)
Answer:
\(2x^2 - 2x + \frac{1}{2} = 0\)
\(4x^2 - 4x + 1 = 0\) ...[Multiplying both sides by 2]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र मध्य पद को विभाजित करने की प्रक्रिया को दर्शाता है। इसमें \(4 \times 1 = 4\) को ऐसे दो गुणनखंडों -2 और -2 में विभाजित किया गया है, जिनका गुणनफल 4 है और योगफल -4 है, जो कि समीकरण में x का गुणांक है।
\(4x^2 - 2x - 2x + 1 = 0\)
\(2x(2x - 1) - 1(2x - 1) = 0\)
\((2x - 1)(2x - 1) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(2x - 1 = 0\) or \(2x - 1 = 0\)
\(2x = 1\) or \(2x = 1\)
\(x = \frac{1}{2}\) or \(x = \frac{1}{2}\)
The roots of the given quadratic equation are \(\frac{1}{2}\) and \(\frac{1}{2}\).
In simple words: First, clear the fraction by multiplying the entire equation by 2, resulting in \(4x^2 - 4x + 1 = 0\). This is a perfect square trinomial, \((2x-1)^2=0\). Solving for x gives repeated roots \(x=\frac{1}{2}\).

🎯 Exam Tip: When dealing with fractional coefficients, clear the fractions first by multiplying by the Least Common Multiple (LCM) of the denominators to simplify the factorization process.

(vi) \(6x - \frac{2}{x} = 1\)
Answer:
\(6x - \frac{2}{x} = 1\)
\(6x^2 - 2 = x\) ...[Multiplying both sides by x]
\(6x^2 - x - 2 = 0\)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र मध्य पद को विभाजित करने की प्रक्रिया को दर्शाता है। इसमें \(6 \times -2 = -12\) को ऐसे दो गुणनखंडों -4 और +3 में विभाजित किया गया है, जिनका गुणनफल -12 है और योगफल -1 है, जो कि समीकरण में x का गुणांक है।
\(6x^2 - 4x + 3x - 2 = 0\)
\(2x(3x - 2) + 1(3x - 2) = 0\)
\((3x - 2)(2x + 1) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(3x - 2 = 0\) or \(2x + 1 = 0\)
\(3x = 2\) or \(2x = -1\)
\(x = \frac{2}{3}\) or \(x = -\frac{1}{2}\)
The roots of the given quadratic equation are \(\frac{2}{3}\) and \(-\frac{1}{2}\).
In simple words: Convert the equation to standard form \(6x^2 - x - 2 = 0\) by multiplying by x. Then, find two numbers whose product is \((6 \times -2) = -12\) and sum is -1 (which are -4 and 3). Factor by grouping to get \((3x-2)(2x+1)=0\), giving roots \(x=\frac{2}{3}\) and \(x=-\frac{1}{2}\).

🎯 Exam Tip: Always check for possible values of x that make the denominator zero in the original equation. In this case, \(x \neq 0\) which is satisfied by the roots found.

(vii) \(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\)
Answer:
\(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र मध्य पद को विभाजित करने की प्रक्रिया को दर्शाता है। इसमें \( \sqrt{2} \times 5\sqrt{2} = 10 \) को ऐसे दो गुणनखंडों 5 और 2 में विभाजित किया गया है, जिनका गुणनफल 10 है और योगफल 7 है, जो कि समीकरण में x का गुणांक है।
\(\sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0\)
\(x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0\)
\((\sqrt{2}x + 5)(x + \sqrt{2}) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\((\sqrt{2}x + 5) = 0\) or \((x + \sqrt{2}) = 0\)
\(x = -\frac{5}{\sqrt{2}}\) or \(x = -\sqrt{2}\)
\(-\frac{5}{\sqrt{2}}\) and \(-\sqrt{2}\) are roots of the equation.
In simple words: For \(\sqrt{2}x^2+7x+5\sqrt{2}=0\), multiply the first and last coefficients: \(\sqrt{2} \times 5\sqrt{2} = 10\). Find factors of 10 that sum to 7, which are 5 and 2. Split the middle term and factor by grouping to get \((\sqrt{2}x+5)(x+\sqrt{2})=0\), giving roots \(x=-\frac{5}{\sqrt{2}}\) and \(x=-\sqrt{2}\).

🎯 Exam Tip: When dealing with terms involving square roots, remember to simplify product terms like \(\sqrt{2} \times \sqrt{2} = 2\) when finding the product of \(a\) and \(c\).

(viii) \(3x^2 - 2\sqrt{6}x + 2 = 0\)
Answer:
\(3x^2 - 2\sqrt{6}x + 2 = 0\)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र मध्य पद को विभाजित करने की प्रक्रिया को दर्शाता है। इसमें \(3 \times 2 = 6\) को ऐसे दो गुणनखंडों \(-\sqrt{6}\) और \(-\sqrt{6}\) में विभाजित किया गया है, जिनका गुणनफल 6 है और योगफल \(-2\sqrt{6}\) है, जो कि समीकरण में x का गुणांक है।
\(3x^2 - \sqrt{6}x - \sqrt{6}x + 2 = 0\)
\(\sqrt{3}x(\sqrt{3}x - \sqrt{2}) - \sqrt{2}(\sqrt{3}x - \sqrt{2}) = 0\)
\((\sqrt{3}x - \sqrt{2})(\sqrt{3}x - \sqrt{2}) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(\sqrt{3}x - \sqrt{2} = 0\) or \(\sqrt{3}x - \sqrt{2} = 0\)
\(\sqrt{3}x = \sqrt{2}\) or \(\sqrt{3}x = \sqrt{2}\)
\(x = \frac{\sqrt{2}}{\sqrt{3}}\) or \(x = \frac{\sqrt{2}}{\sqrt{3}}\)
The roots of the given quadratic equation are \(\frac{\sqrt{2}}{\sqrt{3}}\) and \(\frac{\sqrt{2}}{\sqrt{3}}\).
In simple words: For \(3x^2 - 2\sqrt{6}x + 2 = 0\), find factors of \((3 \times 2) = 6\) that sum to \(-2\sqrt{6}\). These are \(-\sqrt{6}\) and \(-\sqrt{6}\). Factor by grouping, noting \(3 = \sqrt{3}\sqrt{3}\) and \(2 = \sqrt{2}\sqrt{2}\), to form perfect square \(( \sqrt{3}x - \sqrt{2} )^2 = 0\), leading to repeated roots \(x=\frac{\sqrt{2}}{\sqrt{3}}\).

🎯 Exam Tip: When the middle term involves a square root, try to express the first and last terms as squares of terms involving square roots to identify potential perfect square trinomials.

(ix) 2m(m - 24) = 50
Answer:
\(2m^2 - 48m = 50\)
\(2m^2 - 48m - 50 = 0\)
\(m^2 - 24m - 25 = 0\) ...[Dividing both sides by 2]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र मध्य पद को विभाजित करने की प्रक्रिया को दर्शाता है। इसमें -25 को ऐसे दो गुणनखंडों -25 और +1 में विभाजित किया गया है, जिनका गुणनफल -25 है और योगफल -24 है, जो कि समीकरण में m का गुणांक है।
\(m^2 - 25m + m - 25 = 0\)
\(m(m - 25) + 1(m - 25) = 0\)
\((m - 25)(m + 1) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(m - 25 = 0\) or \(m + 1 = 0\)
\(m = 25\) or \(m = -1\)
The roots of the given quadratic equation are 25 and -1.
In simple words: First, expand and rearrange the equation to \(2m^2 - 48m - 50 = 0\). Divide by 2 to simplify to \(m^2 - 24m - 25 = 0\). Find two numbers that multiply to -25 and add to -24 (which are -25 and 1). Factor to \((m-25)(m+1)=0\), giving roots \(m=25\) and \(m=-1\).

🎯 Exam Tip: Always simplify the quadratic equation by dividing by a common factor if possible, as it makes the factorization much easier and reduces the chance of errors.

(x) 25m² = 9
Answer:
\(25m^2 - 9 = 0\)
\((5m)^2 - (3)^2 = 0\)
\((5m + 3)(5m - 3) = 0\)
[:\((a)^2 - (b)^2 = (a + b)(a - b)\)]
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(5m + 3 = 0\) or \(5m - 3 = 0\)
\(5m = -3\) or \(5m = 3\)
\(m = -\frac{3}{5}\) or \(m = \frac{3}{5}\)
The roots of the given quadratic equation are \(-\frac{3}{5}\) and \(\frac{3}{5}\).
In simple words: Rearrange the equation to \(25m^2 - 9 = 0\), which is in the form of a difference of squares \((5m)^2 - (3)^2 = 0\). Factor it as \((5m+3)(5m-3)=0\), which yields roots \(m=-\frac{3}{5}\) and \(m=\frac{3}{5}\).

🎯 Exam Tip: Recognize quadratic equations that are a difference of squares \( (a^2 - b^2) \) as they can be quickly factored into \( (a-b)(a+b) \), saving time and effort.

(xi) 7m² = 21m
Answer:
\(7m^2 - 21m = 0\)
\(m^2 - 3m = 0\) ...[Dividing both sides by 7]
\(m(m - 3) = 0\)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(m = 0\) or \(m - 3 = 0\)
\(m = 0\) or \(m = 3\)
The roots of the given quadratic equation are 0 and 3.
In simple words: Bring all terms to one side to get \(7m^2 - 21m = 0\). Divide by 7 to simplify to \(m^2 - 3m = 0\). Factor out the common term 'm' to get \(m(m-3)=0\), giving roots \(m=0\) and \(m=3\).

🎯 Exam Tip: When an equation lacks a constant term (\(c=0\)), always factor out the common variable term. One of the roots will always be zero in such cases.

(xii) m² - 11 = 0
Answer:
\(m^2 - (\sqrt{11})^2 = 0\)
\((m + \sqrt{11})(m - \sqrt{11}) = 0\)
...[: \((a)^2 - (b)^2 = (a + b)(a - b)\)]
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\(m + \sqrt{11} = 0\) or \(m - \sqrt{11} = 0\)
\(m = -\sqrt{11}\) or \(m = \sqrt{11}\)
The roots of the given quadratic equation are \(-\sqrt{11}\) and \(\sqrt{11}\).
In simple words: The equation \(m^2 - 11 = 0\) can be seen as a difference of squares: \(m^2 - (\sqrt{11})^2 = 0\). Factor this into \((m+\sqrt{11})(m-\sqrt{11})=0\), which yields roots \(m=-\sqrt{11}\) and \(m=\sqrt{11}\).

🎯 Exam Tip: Any positive constant can be expressed as a square of its square root, allowing the difference of squares factorization to be applied even when a perfect square number is not immediately obvious.