Question

Solve by completing the square method

Answer 1

$\underline {\pink { Completing \:Square \: Method : }}$

$Given \: Quadratic \: Equation :\\4x^{2} - \sqrt{3}x - 5 = 0$

$\implies 4x^{2} - \sqrt{3}x = 5$

/* Dividing each term by 4,we get */

$\implies x^{2} - \frac{\sqrt{3}}{4} x = \frac{5}{4}$

$\implies x^{2} - 2 \times \frac{\sqrt{3}}{8} x = \frac{5}{4}$

$\implies x^{2} - 2 \times \frac{\sqrt{3}}{8} x + \Big(\frac{\sqrt{3}}{8}\Big)^{2} = \frac{5}{4}+ \Big(\frac{\sqrt{3}}{8}\Big)^{2}$

$\implies \Big ( x - \frac{\sqrt{3}}{8} )\Big)^{2} = \frac{5}{4} + \frac{3}{64}$

$\implies \Big ( x - \frac{\sqrt{3}}{8} )\Big)^{2} = \frac{80 + 3}{64}$

$\implies \Big ( x - \frac{\sqrt{3}}{8} )\Big)^{2} = \frac{83}{64}$

$\implies \Big ( x - \frac{\sqrt{3}}{8} )\Big)^{2} = \pm \sqrt{ \frac{83}{64}}$

$\implies x - \frac{\sqrt{3}}{8} = \pm \sqrt{ \frac{83}{64}}$

$\implies x = \frac{\sqrt{3}}{8} \pm \frac{\sqrt{83}}{8}$

$\implies x = \frac{\sqrt{3}\pm \sqrt{83}}{8}$

Therefore.,

$\green { x = \frac{\sqrt{3}+\sqrt{83}}{8} \:Or \: </p><p>\frac{\sqrt{3}- \sqrt{83}}{8} }$

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Answer 2

Answer:

Step-by-step explanation: