# Solve the following

Question:

$8 x^{2}-9 x+3=0$

Solution:

Given: $8 x^{2}-9 x+3=0$

Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=8, b=-9$ and $c=3$.

Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get:

$\alpha=\frac{9+\sqrt{81-4 \times 8 \times 3}}{2 \times 8}$ and $\beta=\frac{9-\sqrt{81-4 \times 8 \times 3}}{2 \times 8}$

$\Rightarrow \alpha=\frac{9+\sqrt{81-96}}{16} \quad$ and $\quad \beta=\frac{9-\sqrt{81-96}}{16}$

$\Rightarrow \alpha=\frac{9+\sqrt{-15}}{16} \quad$ and $\quad \beta=\frac{9-\sqrt{-15}}{16}$

$\Rightarrow \alpha=\frac{9+\sqrt{15 i^{2}}}{16} \quad$ and $\quad \beta=\frac{9-\sqrt{15 i^{2}}}{16}$

$\Rightarrow \alpha=\frac{9+i \sqrt{15}}{16} \quad$ and $\quad \beta=\frac{9-i \sqrt{15}}{16}$

$\Rightarrow \alpha=\frac{9}{16}-\frac{\sqrt{15}}{16} i$ and $\beta=\frac{9}{16}+\frac{\sqrt{15}}{16} i$

Hence, the roots of the equation are $\frac{9}{16} \pm \frac{\sqrt{15}}{16} i$.