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Quadratic Equations — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsQuadratic Equations3 Marks
Question
In the following, determine whether the given values are solution of the given equation or not:
$\text{x}^2-3\sqrt{3}\text{x}+6=0,$ $\text{x}=\sqrt{3},\text{x}=-2\sqrt{3}$
✓

Answer

We have been given that,
$\text{x}^2-3\sqrt{3}\text{x}+6=0,$ $\text{x}=\sqrt{3},\text{x}=-2\sqrt{3}$
Now, if $\text{x}=\sqrt{3}$ is a solution then it should satisfy the equation so, substituting $\text{x}=\sqrt{3}$ in the equation, we get
$\text{x}^2-3\sqrt{3}\text{x}+6=(\sqrt{3})^2-3\sqrt{3}(\sqrt{3})+6$
$=3-9+6$
$=0$
$\text{x}=-2\sqrt{3}$
$(-2\sqrt{3})^2-3\sqrt{3}(-2\sqrt{3})+6$
$4\times3+6\times3+6$
$=12+18+6$
$=36$
Which in not equal to zero.
Hence, $\text{x}=-2\sqrt{3}$ is a solution of the quadratic equation.
Therefore, from the above results we find out that $\text{x}=\sqrt{3}$ and $\text{x}=-2\sqrt{3}$ is not solutions of the given quadratic equation.

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