Question

$21x^2-28x+10=0$

Answer 1

Given: $21x^2-28x+10=0$
Comparing the given equation with the general form of the quadratic equation $a{x}^2+bx+c=0$, we get $a=21,b=-28$ and $c=10$.


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

$\alpha =\frac{28+\sqrt{784-4\times 21\times 10}}{2\times 21}$ and $\beta =\frac{28-\sqrt{784-4\times 21\times 10}}{2\times 21}$

$\Rightarrow \alpha =\frac{28+\sqrt{-56}}{42}$ and $\beta =\frac{28-\sqrt{-56}}{42}$

$\Rightarrow \alpha =\frac{28+2i\sqrt{14}}{42}$ and $\beta =\frac{28-2i\sqrt{14}}{42}$

$\Rightarrow \alpha =\frac{14+i\sqrt{14}}{21}$ and $\beta =\frac{14-i\sqrt{14}}{21}$

$\Rightarrow \alpha =\frac{2}{3}+\frac{\sqrt{14}}{21}i$ and $\beta =\frac{2}{3}-\frac{\sqrt{14}}{21}i$



Hence, the roots of the equation are $\frac{2}{3}\pm \frac{\sqrt{14}}{21}$.