In the following, determine whether the given values are solutions of the given equation or not:
$x^2\ –\ 3x\ +\ 2\ =\ 0,\ x\ =\ 2,\ x\ =\ –1$

Given:

The given equation is $x^2\ –\ 3x\ +\ 2\ =\ 0$.

To do:

We have to determine whether $x=2, x=-1$ are solutions of the given equation.

Solution:

If the given values are the solutions of the given equation then they should satisfy the given equation.

Therefore,

For $x=2$,

LHS$=x^2-3x+2$

        $=(2)^2-3(2)+2$

        $=4-6+2$

       $=0$

       $=$RHS

Hence, $x=2$ is a solution of the given equation.

For $x=-1$,

LHS$=x^2-3x+2$

        $=(-1)^2-3(-1)+2$

        $=1+3+2$

        $=6$

RHS$=0$

LHS$≠$RHS

Hence, $x=-1$ is not a solution of the given equation.