Maharashtra Board Class 10 Maths Chapter 2 Quadratic Equations Set 2.3 Solutions

Question 1. Solve the following quadratic equations by completing the square method.

Question 1. x² + x − 20 = 0
Answer: If \(x^2 + x + k = (x + a)^2\), then
\(x^2 + x + k = x^2 + 2ax + a^2\)
Comparing the coefficients, we get
\(1 = 2a\) and \(k = a^2\)

\( \implies a = \frac{1}{2}\) and \(k = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\)
Now, \(x^2 + x - 20 = 0\)

\( \implies x^2 + x + \frac{1}{4} - \frac{1}{4} - 20 = 0\)

\( \implies \left(x + \frac{1}{2}\right)^2 - \left(\frac{1 + 80}{4}\right) = 0\)

\( \implies \left(x + \frac{1}{2}\right)^2 - \frac{81}{4} = 0\)

\( \implies \left(x + \frac{1}{2}\right)^2 = \frac{81}{4}\)
Taking square root of both sides, we get
\(x + \frac{1}{2} = \pm \frac{9}{2}\)

\( \implies x + \frac{1}{2} = \frac{9}{2}\) or \(x + \frac{1}{2} = -\frac{9}{2}\)

\( \implies x = \frac{9}{2} - \frac{1}{2}\) or \(x = -\frac{9}{2} - \frac{1}{2}\)

\( \implies x = \frac{8}{2}\) or \(x = -\frac{10}{2}\)

\( \implies x = 4\) or \(x = -5\)
The roots of the given quadratic equation are 4 and -5.
In simple words: To solve \(x^2 + x - 20 = 0\) by completing the square, we first find a constant to make the left side a perfect square trinomial. This involves comparing it to \((x+a)^2\) to find \(a\) and \(k\). Then we rewrite the equation, isolate the perfect square, and take the square root of both sides to find the values of \(x\).

🎯 Exam Tip: Remember to add and subtract the same term to complete the square, ensuring the equation remains balanced. Pay close attention to signs when taking the square root and solving for \(x\). This method is crucial for understanding the derivation of the quadratic formula.

Question 2. x² + 2x − 5 = 0
Answer: If \(x^2 + 2x + k = (x + a)^2\), then
\(x^2 + 2x + k = x^2 + 2ax + a^2\)
Comparing the coefficients, we get
\(2 = 2a\) and \(k = a^2\)

\( \implies a = 1\) and \(k = (1)^2 = 1\)
Now, \(x^2 + 2x - 5 = 0\)

\( \implies x^2 + 2x + 1 - 1 - 5 = 0\)

\( \implies (x + 1)^2 - 6 = 0\)

\( \implies (x + 1)^2 = 6\)
Taking square root of both sides, we get
\(x + 1 = \pm \sqrt{6}\)

\( \implies x + 1 = \sqrt{6}\) or \(x + 1 = -\sqrt{6}\)

\( \implies x = \sqrt{6} - 1\) or \(x = -\sqrt{6} - 1\)
The roots of the given quadratic equation are \(\sqrt{6} - 1\) and \(-\sqrt{6} - 1\).
In simple words: To solve \(x^2 + 2x - 5 = 0\) using the completing the square method, we identify the constant needed to form a perfect square, add and subtract it, then factorize. By isolating the squared term and taking the square root, we can solve for \(x\).

🎯 Exam Tip: When the constant term is not easily manageable for a perfect square, you'll often end up with irrational roots involving square roots. Ensure you handle the \(\pm\) sign correctly when taking square roots to get both possible solutions.

Question 3. m² - 5m = -3
Answer: First, rearrange the equation into standard form:
\(m^2 - 5m + 3 = 0\)
If \(m^2 - 5m + k = (m + a)^2\), then
\(m^2 - 5m + k = m^2 + 2am + a^2\)
Comparing the coefficients, we get
\(-5 = 2a\) and \(k = a^2\)

\( \implies a = -\frac{5}{2}\) and \(k = \left(-\frac{5}{2}\right)^2 = \frac{25}{4}\)
Now, \(m^2 - 5m + 3 = 0\)

\( \implies m^2 - 5m + \frac{25}{4} - \frac{25}{4} + 3 = 0\)

\( \implies \left(m - \frac{5}{2}\right)^2 + \left(\frac{-25 + 12}{4}\right) = 0\)

\( \implies \left(m - \frac{5}{2}\right)^2 - \frac{13}{4} = 0\)

\( \implies \left(m - \frac{5}{2}\right)^2 = \frac{13}{4}\)
Taking square root of both sides, we get
\(m - \frac{5}{2} = \pm \frac{\sqrt{13}}{2}\)

\( \implies m - \frac{5}{2} = \frac{\sqrt{13}}{2}\) or \(m - \frac{5}{2} = -\frac{\sqrt{13}}{2}\)

\( \implies m = \frac{\sqrt{13}}{2} + \frac{5}{2}\) or \(m = -\frac{\sqrt{13}}{2} + \frac{5}{2}\)

\( \implies m = \frac{\sqrt{13} + 5}{2}\) or \(m = \frac{5 - \sqrt{13}}{2}\)
The roots of the given quadratic equation are \(\frac{\sqrt{13} + 5}{2}\) and \(\frac{5 - \sqrt{13}}{2}\).
In simple words: To solve \(m^2 - 5m = -3\), first rearrange it into standard quadratic form \(m^2 - 5m + 3 = 0\). Then, identify the term needed to complete the square by comparing coefficients, add and subtract it, and solve for \(m\) by taking the square root.

🎯 Exam Tip: Fractions often appear when completing the square, especially when the middle term's coefficient is odd. Be careful with arithmetic operations involving fractions and square roots to avoid common errors.

Question 4. 9y² - 12y + 2 = 0
Answer: Divide both sides by 9 to make the leading coefficient 1:
\( \implies y^2 - \frac{12}{9}y + \frac{2}{9} = 0\)
\( \implies y^2 - \frac{4}{3}y + \frac{2}{9} = 0\)
If \(y^2 - \frac{4}{3}y + k = (y + a)^2\), then
\(y^2 - \frac{4}{3}y + k = y^2 + 2ay + a^2\)
Comparing the coefficients, we get
\(-\frac{4}{3} = 2a\) and \(k = a^2\)

\( \implies a = -\frac{4}{3} \times \frac{1}{2} = -\frac{2}{3}\) and \(k = \left(-\frac{2}{3}\right)^2 = \frac{4}{9}\)
Now, \(y^2 - \frac{4}{3}y + \frac{2}{9} = 0\)

\( \implies y^2 - \frac{4}{3}y + \frac{4}{9} - \frac{4}{9} + \frac{2}{9} = 0\)

\( \implies \left(y - \frac{2}{3}\right)^2 - \frac{2}{9} = 0\)

\( \implies \left(y - \frac{2}{3}\right)^2 = \frac{2}{9}\)
Taking square root of both sides, we get
\(y - \frac{2}{3} = \pm \frac{\sqrt{2}}{3}\)

\( \implies y - \frac{2}{3} = \frac{\sqrt{2}}{3}\) or \(y - \frac{2}{3} = -\frac{\sqrt{2}}{3}\)

\( \implies y = \frac{\sqrt{2}}{3} + \frac{2}{3}\) or \(y = -\frac{\sqrt{2}}{3} + \frac{2}{3}\)

\( \implies y = \frac{\sqrt{2} + 2}{3}\) or \(y = \frac{-\sqrt{2} + 2}{3}\)
The roots of the given quadratic equation are \(\frac{\sqrt{2} + 2}{3}\) and \(\frac{-\sqrt{2} + 2}{3}\).
In simple words: When the leading coefficient is not 1, divide the entire equation by that coefficient first to simplify. Then proceed with completing the square as usual by finding the constant, adding and subtracting it, and solving for the variable.

🎯 Exam Tip: Always make sure the coefficient of the squared term is 1 before starting to complete the square. Forgetting this step is a common mistake and will lead to incorrect results. Simplifying fractions is also important.

Question 5. 2y² + 9y + 10 = 0
Answer: Divide both sides by 2:
\( \implies y^2 + \frac{9}{2}y + 5 = 0\)
If \(y^2 + \frac{9}{2}y + k = (y + a)^2\), then
\(y^2 + \frac{9}{2}y + k = y^2 + 2ay + a^2\)
Comparing the coefficients, we get
\(\frac{9}{2} = 2a\) and \(k = a^2\)

\( \implies a = \frac{9}{2} \times \frac{1}{2} = \frac{9}{4}\) and \(k = \left(\frac{9}{4}\right)^2 = \frac{81}{16}\)
Now, \(y^2 + \frac{9}{2}y + 5 = 0\)

\( \implies y^2 + \frac{9}{2}y + \frac{81}{16} - \frac{81}{16} + 5 = 0\)

\( \implies \left(y + \frac{9}{4}\right)^2 + \left(\frac{-81 + 80}{16}\right) = 0\)

\( \implies \left(y + \frac{9}{4}\right)^2 - \frac{1}{16} = 0\)

\( \implies \left(y + \frac{9}{4}\right)^2 = \frac{1}{16}\)
Taking square root of both sides, we get
\(y + \frac{9}{4} = \pm \frac{1}{4}\)

\( \implies y + \frac{9}{4} = \frac{1}{4}\) or \(y + \frac{9}{4} = -\frac{1}{4}\)

\( \implies y = \frac{1}{4} - \frac{9}{4}\) or \(y = -\frac{1}{4} - \frac{9}{4}\)

\( \implies y = \frac{-8}{4}\) or \(y = \frac{-10}{4}\)

\( \implies y = -2\) or \(y = -\frac{5}{2}\)
The roots of the given quadratic equation are \(-2\) and \(-\frac{5}{2}\).
In simple words: For \(2y^2 + 9y + 10 = 0\), first divide by 2 to make the coefficient of \(y^2\) one. Then, complete the square by adding and subtracting the required constant. Finally, solve for \(y\) by taking the square root.

🎯 Exam Tip: Be meticulous with calculations involving fractions, especially during the common denominator step when combining constants. A small arithmetic error can lead to entirely wrong roots.

Question 6. 5x² = 4x + 7
Answer: First, rearrange into standard form \(ax^2 + bx + c = 0\):
\( \implies 5x^2 - 4x - 7 = 0\)
Now, divide both sides by 5:
\( \implies x^2 - \frac{4}{5}x - \frac{7}{5} = 0\)
If \(x^2 - \frac{4}{5}x + k = (x + a)^2\), then
\(x^2 - \frac{4}{5}x + k = x^2 + 2ax + a^2\)
Comparing the coefficients, we get
\(-\frac{4}{5} = 2a\) and \(k = a^2\)

\( \implies a = -\frac{4}{5} \times \frac{1}{2} = -\frac{2}{5}\) and \(k = \left(-\frac{2}{5}\right)^2 = \frac{4}{25}\)
Now, \(x^2 - \frac{4}{5}x - \frac{7}{5} = 0\)

\( \implies x^2 - \frac{4}{5}x + \frac{4}{25} - \frac{4}{25} - \frac{7}{5} = 0\)

\( \implies \left(x - \frac{2}{5}\right)^2 - \left(\frac{4 + 35}{25}\right) = 0\)

\( \implies \left(x - \frac{2}{5}\right)^2 - \frac{39}{25} = 0\)

\( \implies \left(x - \frac{2}{5}\right)^2 = \frac{39}{25}\)
Taking square root of both sides, we get
\(x - \frac{2}{5} = \pm \frac{\sqrt{39}}{5}\)

\( \implies x - \frac{2}{5} = \frac{\sqrt{39}}{5}\) or \(x - \frac{2}{5} = -\frac{\sqrt{39}}{5}\)

\( \implies x = \frac{2}{5} + \frac{\sqrt{39}}{5}\) or \(x = \frac{2}{5} - \frac{\sqrt{39}}{5}\)

\( \implies x = \frac{2 + \sqrt{39}}{5}\) or \(x = \frac{2 - \sqrt{39}}{5}\)
The roots of the given quadratic equation are \(\frac{2 + \sqrt{39}}{5}\) and \(\frac{2 - \sqrt{39}}{5}\).
In simple words: Begin by rearranging the equation into the standard quadratic form \(ax^2 + bx + c = 0\). Then, divide by the coefficient of \(x^2\) to normalize it, complete the square by finding and applying the correct constant, and solve for \(x\) by taking the square root.

🎯 Exam Tip: Remember to always bring all terms to one side to get the standard form before attempting to complete the square, especially if terms are on opposite sides of the equality. Also, be careful when simplifying square roots of non-perfect squares.