Solve the given quadratic equation by completing the square, $2 x^{2} + 5 x - 3 = 0$

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Solution

The given quadratic equation is $2 x^{2} + 5 x - 3 = 0$ . Multiply the equation by $2$ then we get $4 x^{2} + 10 x - 6 = 0$ .

Add and subtract $\frac{25}{4}$ in the equation $4 x^{2} + 10 x - 6 = 0$ to make a perfect square as shown below:

$2 x^{2} + 5 x - 3 = 0 \Rightarrow 2 x^{2} + 5 x + \frac{25}{4} - \frac{25}{4} - 3 = 0 \Rightarrow (2 x)^{2} + (2 \times 2 x \times \frac{5}{2}) + {(\frac{5}{2})}^{2} + (- \frac{25}{4} - 3) = 0 \Rightarrow {(2 x + \frac{5}{2})}^{2} - \frac{49}{4} = 0 (∵ (a + b)^{2} = a^{2} + b^{2} + 2 a b) \Rightarrow {(2 x + \frac{5}{2})}^{2} = \frac{49}{4} \Rightarrow 2 x + \frac{5}{2} = \pm \sqrt{\frac{49}{4}} \Rightarrow 2 x + \frac{5}{2} = \pm \frac{7}{2} \Rightarrow 2 x + \frac{5}{2} = - \frac{7}{2}, 2 x + \frac{5}{2} = \frac{7}{2} \Rightarrow 2 x = - \frac{7}{2} - \frac{5}{2}, 2 x = \frac{7}{2} - \frac{5}{2} \Rightarrow 2 x = - \frac{12}{2}, 2 x = \frac{2}{2} \Rightarrow 2 x = - 6, 2 x = 1 \Rightarrow x = - \frac{6}{2}, x = \frac{1}{2} \Rightarrow x = - 3, x = \frac{1}{2}$

Hence, $x = \frac{1}{2}$ or $x = - 3$ .

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