Solve this


$x^{2}-6 x+4=0$


Given :

$x^{2}-6 x+4=0$

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

$a=1, b=-6$ and $c=4$

Discriminant $D$ is given by :

$D=\left(b^{2}-4 a c\right)$

$=(-6)^{2}-4 \times 1 \times 4$



Hence, the roots of the equation are real.

Roots $\alpha$ and $\beta$ are given by :

$\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-6)+\sqrt{20}}{2 \times 1}=\frac{6+2 \sqrt{5}}{2}=\frac{2(3+\sqrt{5})}{2}=(3+\sqrt{5})$

$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-6)-\sqrt{20}}{2 \times 1}=\frac{6-2 \sqrt{5}}{2}=\frac{2(3-\sqrt{5})}{2}=(3-\sqrt{5})$

Thus, the roots of the equation are $(3+2 \sqrt{5})$ and $(3-2 \sqrt{5})$.


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