Question

Find the value of $k$ for which the given value is a solution of the given equation

$3\sqrt{7}{x}^2-4x+k=0;x=\frac{\sqrt{7}}{3}$

• A. $-\sqrt{7}$
• B. $\sqrt{7}$
• C. $-\sqrt{5}$
• D. $\sqrt{5}$

Answer 1

Answer: (A)

$-\sqrt{7}$

Given equation,

$3\sqrt{7}{x}^2-4x+k=0$ and $x=\frac{\sqrt{7}}{3}$

$\Rightarrow 3\sqrt{7}(\frac{\sqrt{7}}{3}{)}^2-4(\frac{\sqrt{7}}{3})+k=0$

$\Rightarrow 3\sqrt{7}\times \frac{7}{9}-\frac{4\sqrt{7}}{3}+k=0$

$\Rightarrow \frac{7\sqrt{7}}{3}-\frac{4\sqrt{7}}{3}+k=0$

$\Rightarrow \frac{3\sqrt{7}}{3}+k=0$

$\Rightarrow \sqrt{7}+k=0$

$\Rightarrow k=-\sqrt{7}$