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Quadratic Equations — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsQuadratic Equations4 Marks
Question
In the following, determine whether the given values are solution of the given equation or not:
$\text{x}^2-\sqrt{2}\text{x}-4=0,$ $\text{x}=-\sqrt{2},$ $\text{x}=-2\sqrt{2}$
✓

Answer

$\text{x}^2-\sqrt{2}\text{x}-4=0,$ $\text{x}=-\sqrt{2},$ $\text{x}=-2\sqrt{2}$When, $\text{x}=-\sqrt{2}$
Substituting $\text{x}=-\sqrt{2}$
L.H.S.
$=\text{x}^2-\sqrt{2}\text{x}-4$
$=(-\sqrt{2})^2-\sqrt{2}(-\sqrt{2})-4$
$=2+2-4=0$
= R.H.S.
$\therefore\text{x}=-\sqrt{2}$ is its solution
When, $\text{x}=-2\sqrt{2}$
Substituting $\text{x}=-2\sqrt{2}$
L.H.S.
$=\text{x}^2-\sqrt{2}\text{x}-4$
$=(-2\sqrt{2})^2-\sqrt{2}(-2\sqrt{2})-4$
$=8+4-4$
$=8\neq\text{R.H.S}$
$\therefore\text{x}=-2\sqrt{2}$ is not its solution.

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