Find the roots of each of the following equations, if they exist, by applying the quadratic formula:
Question:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$\sqrt{3} x^{2}-2 \sqrt{2} x-2 \sqrt{3}=0$

 

Solution:

The given equation is $\sqrt{3} x^{2}-2 \sqrt{2} x-2 \sqrt{3}=0$.

Comparing it with $a x^{2}+b x+c=0$, we get

$a=\sqrt{3}, b=-2 \sqrt{2}$ and $c=-2 \sqrt{3}$

$\therefore$ Discriminant, $D=b^{2}-4 a c=(-2 \sqrt{2})^{2}-4 \times \sqrt{3} \times(-2 \sqrt{3})=8+24=32>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{32}=4 \sqrt{2}$

$\therefore \alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-2 \sqrt{2})+4 \sqrt{2}}{2 \times \sqrt{3}}=\frac{6 \sqrt{2}}{2 \sqrt{3}}=\sqrt{6}$

$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-2 \sqrt{2})-4 \sqrt{2}}{2 \times \sqrt{3}}=\frac{-2 \sqrt{2}}{2 \sqrt{3}}=-\frac{\sqrt{6}}{3}$

Hence, $\sqrt{6}$ and $-\frac{\sqrt{6}}{3}$ are the roots of the given equation.

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