If the roots of the quadratic equation 5x(2x | vedclass

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Question

(D) Given equation: $5x(2x - 3) - 4 = 0$ Expand the equation: $10x^2 - 15x - 4 = 0$ Divide by $10$: $x^2 - \frac{15}{10}x - \frac{4}{10} = 0 \Rightarr

Vedclass · Accepted Answer

$\frac{15+\sqrt{385}}{20}, \frac{15-\sqrt{385}}{20}$

Vedclass · Answer

(D) Given equation: $5x(2x - 3) - 4 = 0$ Expand the equation: $10x^2 - 15x - 4 = 0$ Divide by $10$: $x^2 - \frac{15}{10}x - \frac{4}{10} = 0 \Rightarrow x^2 - \frac{3}{2}x - \frac{2}{5} = 0$ Add and subtract $(\frac{1}{2} 	imes 	ext{coefficient of } x)^2 = (\frac{1}{2} 	imes \frac{3}{2})^2 = (\frac{3}{4})^2 = \frac{9}{16}$ $x^2 - \frac{3}{2}x + \frac{9}{16} - \frac{9}{16} - \frac{2}{5} = 0$ $(x - \frac{3}{4})^2 = \frac{9}{16} + \frac{2}{5} = \frac{45 + 32}{80} = \frac{77}{80}$ $x - \frac{3}{4} = \pm \sqrt{\frac{77}{80}} = \pm \frac{\sqrt{77}}{4\sqrt{5}} = \pm \frac{\sqrt{385}}{20}$ $x = \frac{3}{4} \pm \frac{\sqrt{385}}{20} = \frac{15 \pm \sqrt{385}}{20}$ Thus,the roots are $\frac{15+\sqrt{385}}{20}$ and $\frac{15-\sqrt{385}}{20}$.

If the roots of the quadratic equation $5x(2x - 3) - 4 = 0$ exist,find them using the method of completing the square.

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