Question

Find the roots of following quadratic equation

$x^2+\frac{x-1}{3}=0$

• A. $x=\frac{-1\pm \sqrt{13}}{6}$
• B. $x=\frac{1\pm \sqrt{13}}{6}$
• C. $x=\frac{-1\pm \sqrt{11}}{6}$
• D. $x=\frac{-1\pm \sqrt{19}}{6}$

Answer 1

Answer: (A)

$x=\frac{-1\pm \sqrt{13}}{6}$

Given equation is

$x^2+\frac{x-1}{3}=0$
$3x^2+x-1=0$
Using quadratic formula,
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$a=3,b=1,c=-1$

$x=\frac{-1\pm \sqrt{1-4(-1)\left(3\right)}}{2\times 3}$
$\therefore x=\frac{-1\pm \sqrt{13}}{6}$

$\therefore$ the roots of the given equation are $\frac{-1\pm \sqrt{13}}{6}$