Question

If $x=\frac{1}{2\sqrt{3}}$ is a zero of a polynomial $2k{x}^2-7x+k$, find $k.$

Answer 1

$x=\frac{1}{2\sqrt{3}}$ is zero of polynomial $2k{x}^2-7x+k$

$\therefore 2k{\left(\frac{1}{2\sqrt{3}}\right)}^2-\left(\frac{7}{2\sqrt{3}}\right)+k=0$

$\Rightarrow \frac{2k}{12}-\frac{7}{2\sqrt{3}}+k=0\to \frac{k}{6}-\frac{7}{2\sqrt{3}}+k=0$

$\Rightarrow k(1+\frac{1}{6})=\frac{7}{2\sqrt{3}}\to k\left(\frac{7}{6}\right)=\frac{7}{2\sqrt{3}}$

$\Rightarrow k=\frac{6}{2\sqrt{3}}=\sqrt{3}$

$\Rightarrow k=\sqrt{3}.$

Hence, the answer is $\sqrt{3}.$