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In the following, determine whether the given values are solutions of the given equation or not:
$x^2\ β\ 3x\ +\ 2\ =\ 0,\ x\ =\ 2,\ x\ =\ β1$
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.
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