# Q1.    Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:                (ii)    $3x^2 - 4\sqrt3x + 4 = 0$

S safeer

$b^2-4ac=(-4\sqrt{3})^2-(4\times4\times3)=48-48=0$

Here the value of discriminant =0, which implies that roots exist and the roots are equal.

The roots are given by the formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\sqrt{3}\pm\sqrt{0}}{2\times3}=\frac{2}{\sqrt{3}}$

So the roots are

$\frac{2}{\sqrt{3}},\ \frac{2}{\sqrt{3}}$

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