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Question
Find the roots of the following quadratic equation by the method of completing the square:
\( 2x^2 + x - 4 = 0\)
\( 2x^2 + x - 4 = 0\)
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Solution
\( 2x^2 + x = 4\)
On dividing both sides of the equation by 2, we get
\( x^2 + \frac{x}{2} = 2\)
On adding the square of one-half of the coefficient of x to both sides [i.e., \((\frac{1}{4})^2\)] to both sides of the equation, we get,
\((x)^2 + \frac{x}{2} + (\frac{1}{4})^2 = 2 + (\frac{1}{4})^2\)
\(\Rightarrow (x)^2 + 2 \times x \times \frac{1}{4} + (\frac{1}{4})^2 = 2 + (\frac{1}{4})^2\\
\Rightarrow (x+\frac{1}{4})^2 = \frac{33}{16}\\
\Rightarrow x + \frac{1}{4} = \pm \frac{\sqrt{33}}{4}\)
\(\Rightarrow x\) = \(\pm \frac{\sqrt{33}}{4} - \frac{1}{4}\)
\(\Rightarrow x\) = \(\frac{\pm \sqrt{33}-1}{4} \)
\(\Rightarrow x\)= \(\frac{\sqrt{33}-1}{4} \) or \(x = \frac{-\sqrt{33}-1}{4} \)
On dividing both sides of the equation by 2, we get
\( x^2 + \frac{x}{2} = 2\)
On adding the square of one-half of the coefficient of x to both sides [i.e., \((\frac{1}{4})^2\)] to both sides of the equation, we get,
\((x)^2 + \frac{x}{2} + (\frac{1}{4})^2 = 2 + (\frac{1}{4})^2\)
\(\Rightarrow (x)^2 + 2 \times x \times \frac{1}{4} + (\frac{1}{4})^2 = 2 + (\frac{1}{4})^2\\
\Rightarrow (x+\frac{1}{4})^2 = \frac{33}{16}\\
\Rightarrow x + \frac{1}{4} = \pm \frac{\sqrt{33}}{4}\)
\(\Rightarrow x\) = \(\pm \frac{\sqrt{33}}{4} - \frac{1}{4}\)
\(\Rightarrow x\) = \(\frac{\pm \sqrt{33}-1}{4} \)
\(\Rightarrow x\)= \(\frac{\sqrt{33}-1}{4} \) or \(x = \frac{-\sqrt{33}-1}{4} \)
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