Question
In the following, determine whether the given values are solution of the given equation or not:
$x^2 - 3x + 2 = 0, x = 2, x = -1$
$x^2 - 3x + 2 = 0, x = 2, x = -1$
✓
Answer
$x^2-3 x+2=0 . x=2, x=-1$
Here, L.H.S. $=x^2-3 x+2$ and R.H.S. $=0$
Now, substitute $x=2$ in LH.S.
$\text { We get }(2)^2-3(2)+2=4-6+2$
$=6-6$
$=0$
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 $(-1)^2-3(-1)+2$
$1+3+2=6 \neq \text { R.H.S. }$
Since L.H.S $\neq$ R.H.S.
$x=-1$ is not a solution for thr given equatoion.
Here, L.H.S. $=x^2-3 x+2$ and R.H.S. $=0$
Now, substitute $x=2$ in LH.S.
$\text { We get }(2)^2-3(2)+2=4-6+2$
$=6-6$
$=0$
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 $(-1)^2-3(-1)+2$
$1+3+2=6 \neq \text { R.H.S. }$
Since L.H.S $\neq$ R.H.S.
$x=-1$ is not a solution for thr given equatoion.
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