Question

Discuss the nature of roots of the given equation: $2n^2+5n-1=0$

Answer 1

We know that while finding the root of a quadratic equation $a{x}^2+bx+c=0$ by quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$,

if $b^2-4ac>0$, then the roots are real and distinct

if $b^2-4ac=0$, then the roots are real and equal and

if $b^2-4ac<0$, then the roots are imaginary.

Here, the given quadratic equation $2n^2+5n-1=0$ is in the form $a{x}^2+bx+c=0$ where $a=2,b=5$ and $c=-1$, therefore,

$b^2-4ac=(5{)}^2-(4\times 2\times -1)=25+8=33>0$

Since $b^2-4ac>0$

Hence the roots are real and distinct.