If the roots of the quadratic equation 3x^{2} | vedclass

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Question

(C) Given equation: $3x^{2} + 2\sqrt{5}x - 5 = 0$.Divide the entire equation by $3$ to make the coefficient of $x^{2}$ equal to $1$:$x^{2} + \frac{2\s

Vedclass · Accepted Answer

$-\sqrt{5}, \frac{\sqrt{5}}{3}$

Vedclass · Answer

(C) Given equation: $3x^{2} + 2\sqrt{5}x - 5 = 0$.Divide the entire equation by $3$ to make the coefficient of $x^{2}$ equal to $1$:$x^{2} + \frac{2\sqrt{5}}{3}x - \frac{5}{3} = 0$.Move the constant term to the right side:$x^{2} + \frac{2\sqrt{5}}{3}x = \frac{5}{3}$.Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $\frac{2\sqrt{5}}{3}$,so half of it is $\frac{\sqrt{5}}{3}$. The square is $(\frac{\sqrt{5}}{3})^{2} = \frac{5}{9}$.$x^{2} + \frac{2\sqrt{5}}{3}x + \frac{5}{9} = \frac{5}{3} + \frac{5}{9}$.$(x + \frac{\sqrt{5}}{3})^{2} = \frac{15 + 5}{9} = \frac{20}{9}$.Taking the square root on both sides:$x + \frac{\sqrt{5}}{3} = \pm \sqrt{\frac{20}{9}} = \pm \frac{2\sqrt{5}}{3}$.Case $1$: $x = \frac{2\sqrt{5}}{3} - \frac{\sqrt{5}}{3} = \frac{\sqrt{5}}{3}$.Case $2$: $x = -\frac{2\sqrt{5}}{3} - \frac{\sqrt{5}}{3} = -\frac{3\sqrt{5}}{3} = -\sqrt{5}$.Thus,the roots are $-\sqrt{5}$ and $\frac{\sqrt{5}}{3}$.

If the roots of the quadratic equation $3x^{2} + 2\sqrt{5}x - 5 = 0$ exist,find them using the method of completing the square.

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