In the following, determine whether the given values are solutions of the given equation or not:
$x^2\ β\ \sqrt{2}x\ β\ 4\ =\ 0,\ x\ =\ -\sqrt{2},\ x\ =\ -2\sqrt{2}$
Given:
The given equation is $x^2\ –\ \sqrt{2}x\ –\ 4\ =\ 0$.
To do:
We have to determine whether $x=-\sqrt{2}, x=-2\sqrt{2}$ 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 $x=-\sqrt{2}$,
LHS$=x^2 β \sqrt{2}x-4$
$=(-\sqrt{2})^2-\sqrt{2}(-\sqrt{2})-4$
$=2+2-4$
$=0$
$=$RHS
Hence, $x=-\sqrt{2}$ is a solution of the given equation.
For $x=-2\sqrt{2}$,
LHS$=x^2 β \sqrt{2}x - 4$
$=(-2\sqrt{2})^2-\sqrt{2}(-2\sqrt{2})-4$
$=8+4-4$
$=8$
RHS$=0$
LHS$≠$RHS
Hence, $x=-2\sqrt{2}$ is not a solution of the given equation.ββ
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