Question

Determine whether the given values of variable is a solution of the quadratic equation or not. $6x^2-x-2=0;x=-\frac{1}{2}$ and $x=\frac{2}{3}$

Answer 1

The given quadratic equation is $6x^2-x-2$.

Substitute $x=-\frac{1}{2}$ in the equation $6x^2-x-2$ as follows:

$6x^2-x-2=0\Rightarrow 6{(-\frac{1}{2})}^2-(-\frac{1}{2})-2=0\Rightarrow (6\times \frac{1}{4})+\frac{1}{2}-2=0\Rightarrow \frac{3}{2}+\frac{1}{2}-2=0\Rightarrow \frac{4}{2}-2=0\Rightarrow 2-2=0\Rightarrow 0=0$

which is true.

Therefore, $x=-\frac{1}{2}$ is the solution of the given quadratic equation.

Now, substitute $x=\frac{2}{3}$ in the equation $6x^2-x-2$ as follows:

$6x^2-x-2=0\Rightarrow 6{\left(\frac{2}{3}\right)}^2-\left(\frac{2}{3}\right)-2=0\Rightarrow (6\times \frac{4}{9})-\frac{2}{3}-2=0\Rightarrow \frac{8}{3}-\frac{2}{3}-2=0\Rightarrow \frac{6}{3}-2=0\Rightarrow 2-2=0\Rightarrow 0=0$

which is true.

Therefore, $x=\frac{2}{3}$ is not the solution of the given quadratic equation.

Hence, $x=-\frac{1}{2}$ and $x=\frac{2}{3}$ are the solution of the quadratic equation $6x^2-x-2$.