Question

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

$2x^2-2\sqrt{2}x+1=0$

Answer 1

The given equation is $2x^2-2\sqrt{2}x+1=0$.

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

a = 2, b = $-2\sqrt{2}$ and c = 1

∴ Discriminant, D = $b^2-4ac={\left(-2\sqrt{2}\right)}^2-4\times 2\times 1=8-8=0$

So, the given equation has real roots.

Now, $\sqrt{D}=0$

$$\begin{gathered} \therefore \alpha =\frac{-b+\sqrt{D}}{2a}=\frac{-\left(-2\sqrt{2}\right)+\sqrt{0}}{2\times 2}=\frac{2\sqrt{2}}{4}=\frac{\sqrt{2}}{2} \\ \beta =\frac{-b-\sqrt{D}}{2a}=\frac{-\left(-2\sqrt{2}\right)-\sqrt{0}}{2\times 2}=\frac{2\sqrt{2}}{4}=\frac{\sqrt{2}}{2} \end{gathered}$$

Hence, $\frac{\sqrt{2}}{2}$ is the repeated root of the given equation.