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

Asked by Abhisek | 1 year ago |  124

Solution :-

3x2 – $$4 \sqrt{3}x$$ + 4 = 0

Comparing the equation with ax2 + bx + c = 0, we get

a = 3, b = $$-4\sqrt{3}$$ and c = 4

We know, Discriminant = b2 – 4ac

= $$( -4\sqrt{3})^2$$ – 4(3)(4)

= 48 – 48 = 0

As b2 – 4ac = 0,

Real roots exist for the given equation and they are equal to each other.

Hence the roots will be $$\dfrac{-b}{2a}\;and\;\dfrac{-b}{2a}$$

$$\dfrac{-b}{2a}=\dfrac{(-4\sqrt{3})}{2\times 3}$$

$$\dfrac{(-4\sqrt{3})}{6}=\dfrac{(2\sqrt{3})}{3}=\dfrac{2}{\sqrt{3}}$$

Therefore, the roots are $$\dfrac{2}{\sqrt{3}}\;and\;\dfrac{2}{\sqrt{3}}$$

Answered by Pragya Singh | 1 year ago

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