Question

If m and n are the roots of the quadratic equation $x^2+6x+2=0$ then $(\frac{1}{m}+\frac{1}{n})$ is ________________.

• A. 12
• B. 3
• C. -12
• D. -3

Answer 1

The correct option is D -3

$x^2+6x+2$
For zeroes,
$a=1,b=6,c=2$
$x=\frac{-6\pm \sqrt{6^2-4\cdot 1\cdot 2}}{2\cdot 1}$
$=\frac{-6\pm \sqrt{36-8}}{2}$
$=\frac{-6\pm \sqrt{28}}{2}$
$=\frac{-6\pm 2\sqrt{7}}{2}$
$=-3\pm \sqrt{7}$
Let, $m=-3+\sqrt{7}$ and $n=-3-\sqrt{7}$
Now,
$\frac{1}{m}+\frac{1}{n}$
$=\frac{1}{-3+\sqrt{7}}+\frac{1}{-3-\sqrt{7}}$
$=\frac{-3-\sqrt{7}+(-3+\sqrt{7})}{(-3+\sqrt{7})(-3-\sqrt{7})}$
$=\frac{-3-3}{(-3{)}^2-(\sqrt{7}{)}^2}$
$=\frac{-6}{9-7}$
$=\frac{-6}{2}=-3$