Question

Solve:

$21x^2=4(2x+1)$

Answer 1

Given quadratic equation is $21x^2=4(2x+1)$

We can simplify the equation to bring it to the standard form.

$21x^2=8x+4$

$21x^2-8x-4=0$

We know that, for a quadratic equation of the form $a{x}^2+bx+c=0$,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, $a=21,b=-8,c=-4$

Hence, $x=\frac{-(-8)\pm \sqrt{{(-8)}^2-4\times 21(-4)}}{2\times 21}$

$=\frac{8\pm \sqrt{64+336}}{42}$

$=\frac{8\pm \sqrt{400}}{42}$

$=\frac{8\pm 20}{42}$

Hence, $x=\frac{28}{42}$ or $\frac{-12}{42}$

$x=\frac{2}{3},-\frac{2}{7}$

Hence, the solution of the given quadratic equation is $x=\frac{2}{3},-\frac{2}{7}$.