In the following, determine whether the given values are the solution of the given equation or not.
$x^{2} + \sqrt{2} x - 4 = 0, x = \sqrt{2}, x = - 2 \sqrt{2}$

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Solution

Substituting $x = \sqrt{2}$ on the LHS of the given equation, we get

LHS $= {(\sqrt{2})}^{2} + \sqrt{2} \times \sqrt{2} - 4 = 2 + 2 - 4 = 0 =$ RHS

So, $x = \sqrt{2}$ is a solution of the given equation

Substituting $x = - 2 \sqrt{2}$ on LHS of the givne equation, we get

LHS $= {(- 2 \sqrt{2})}^{2} + \sqrt{2} \times (- 2 \sqrt{2}) - 4 = 8 - 4 - 4 = 0 =$ RHS

So, $x = - 2 \sqrt{2}$ is also a solution of the given equation.

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In the following, determine whether the given values are the solution of the given equation or not.x2+√2x−4=0,x=√2,x=−2√2

In the following, determine whether the given values are the solution of the given equation or not.
$x^{2} + \sqrt{2} x - 4 = 0, x = \sqrt{2}, x = - 2 \sqrt{2}$