Question

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

$x^2-\left(\sqrt{3}+1\right)x+\sqrt{3}=0$ [CBSE 2015]

Answer 1

The given equation is $x^2-\left(\sqrt{3}+1\right)x+\sqrt{3}=0$.

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

a = 1, b = $-\left(\sqrt{3}+1\right)$ and c = $\sqrt{3}$

∴ Discriminant, D = $b^2-4ac={\left[-\left(\sqrt{3}+1\right)\right]}^2-4\times 1\times \sqrt{3}=3+1+2\sqrt{3}-4\sqrt{3}=3-2\sqrt{3}+1={\left(\sqrt{3}-1\right)}^2>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{{\left(\sqrt{3}-1\right)}^2}=\sqrt{3}-1$

$$\begin{gathered} \therefore \alpha =\frac{-b+\sqrt{D}}{2a}=\frac{-\left[-\left(\sqrt{3}+1\right)\right]+\left(\sqrt{3}-1\right)}{2\times 1}=\frac{\sqrt{3}+1+\sqrt{3}-1}{2}=\frac{2\sqrt{3}}{2}=\sqrt{3} \\ \beta =\frac{-b-\sqrt{D}}{2a}=\frac{-\left[-\left(\sqrt{3}+1\right)\right]-\left(\sqrt{3}-1\right)}{2\times 1}=\frac{\sqrt{3}+1-\sqrt{3}+1}{2}=\frac{2}{2}=1 \end{gathered}$$

Hence, $\sqrt{3}$ and 1 are the roots of the given equation.