Mathematics

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :
3x² - $4\sqrt{3}x + 4$ = 0

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Comparing 3x² - $4\sqrt{3}x + 4$ = 0 with ax² + bx + c = 0, we get :

a = 3, b = $-4\sqrt{3}$ and c = 4.

Substituting values in b² - 4ac, we get :

$\Rightarrow (-4\sqrt{3})^2 - 4 \times 3 \times 4 \\[1em] \Rightarrow 48 - 48 \\[1em] \Rightarrow 0.$

Since, b² - 4ac = 0.

∴ There are two real and equal roots.

For equal roots,

x = $-\dfrac{b}{2a} = -\dfrac{-4\sqrt{3}}{2 \times 3} = \dfrac{4\sqrt{3}}{6} = \dfrac{2\sqrt{3}}{3}$ .

Hence, equal roots are $\dfrac{2}{\sqrt{3}}, \dfrac{2}{\sqrt{3}}$ .

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