Question

Find the roots of following quadratic equation

$7x+1=6x^2$

• A. $x=\frac{5\pm \sqrt{73}}{12}$
• B. $x=\frac{2\pm \sqrt{73}}{12}$
• C. $x=\frac{3\pm \sqrt{73}}{12}$
• D. $x=\frac{7\pm \sqrt{73}}{12}$

Answer 1

The correct option is D $x=\frac{7\pm \sqrt{73}}{12}$

Given equation is

$7x+1=6x^2$
$6x^2-7x-1=0$
Using quadratic formula,
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$a=6,b=-7,c=-1$

$x=\frac{7\pm \sqrt{49-4(-1)\left(6\right)}}{2\times 6}$
$\therefore x=\frac{7\pm \sqrt{73}}{12}$

$\therefore$ the roots of the given equation are $\frac{7\pm \sqrt{73}}{12}$