Question
In the following, determine whether the given values are solution of the given equation or not:
$x^2 - 3x + 2 = 0, x = 2, x = -1$

Answer

$x^2 - 3x + 2 = 0. x = 2, x = -1$
$$Here, $L.H.S. = x^2 - 3x + 2$ and $R.H.S. = 0$
Now, substitute $x = 2$ in $L.H.S.$
We get $(2)^2 - 3(2) + 2 = 4 - 6 + 2$
$= 6 - 6$
$= 0$
R.H.S.
Since, L.H.S. = R.H.S.
x = 2 is a solution for the given equation.
Similarly,
Now substitute x = -1 in L.H.S.
We get $(-1)^2 - 3(-1) + 2$
1 + 3 + 2 = 6 $\neq$ R.H.S.
Since L.H.S $\neq$ R.H.S.
x = -1 is not a solution for thr given equatoion.

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