# Q1.    Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:                (iii)    $2x^2 - 6x + 3 = 0$

S safeer

The value of the discriminant

$b^2-4ac=(-6)^2-4\times2\times3=12$

The discriminant > 0. Therefore the given quadratic equation has two distinct real root

roots are

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-6\pm\sqrt{12}}{2\times2}=\frac{3}{2}\pm\frac{\sqrt{3}}{2}$

So the roots are

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

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