Question

In the following determine whether the given values are solution of the given equation or not:
$6x^2-x-2=0,x=\frac{-1}{2},x=\frac{2}{3}$

Answer 1

Substituting $x=-\frac{1}{2}$ in the LHS of the given equation we get
LHS$=6\times {(-\frac{1}{2})}^2-(-\frac{1}{2})-2=\frac{6}{4}+\frac{1}{2}-2=0=$ RHS
So, $x=\frac{1}{2}$ is a solution of the given equation.
for $x=\frac{2}{3}$ we have
LHS $=6\times {\left(\frac{2}{3}\right)}^2-\frac{2}{3}-2=0=$ RHS
So, $x=\frac{2}{3}$ is also a solution of the given equation