Question

Find the discriminant of the equation and the nature of roots. Also find the roots.

$6x^2+x-2=0$

• A. $D=49$, Real and distinct roots: $\frac{1}{5},\frac{-2}{3}$
• B. $D=39$, Real and distinct roots: $\frac{1}{2},\frac{-2}{3}$
• C. $D=49$, Real and distinct roots: $\frac{1}{3},\frac{-7}{3}$
• D. $D=49$, Real and distinct roots: $\frac{1}{2},\frac{-2}{3}$
  

Answer 1

Answer: (D)

$D=49$, Real and distinct roots: $\frac{1}{2},\frac{-2}{3}$

Nature of the roots of a quadratic equation is determined by its discriminant $D=b^2-4ac$

Comparing $6x^2+x-2=0$ with $a{x}^2+bx+c=0$, we get

$a=6,b=1,c=-2$

Therefore, $D=b^2-4ac$
$=1^2-4\times 6\times (-2)$
$=1+48$
$=49$ $>0$

Therefore, the roots are real.

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

$=\frac{-\left(1\right)\pm \sqrt{(1{)}^2-4\times 6\times (-2)}}{2\times 6}$

$=\frac{-1\pm \sqrt{49}}{12}$

$=\frac{-1\pm 7}{12}$

Therefore,

$x=\frac{-1+7}{12}$ and $x=\frac{-1-7}{12}$

$x=\frac{6}{12}$ and $x=\frac{-8}{12}$

$x=\frac{1}{2}$ and $x=\frac{-2}{3}$