Question

For what positive values of '$m$' roots of given equation is equal, distinct, imaginary $x^2-mx+9=0$

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 $x^2-mx+9=0$ is in the form $a{x}^2+bx+c=0$ where $a=1,b=-m$ and $c=9$.

(i) If the roots are equal then $b^2-4ac=0$, therefore,

$b^2-4ac=0\Rightarrow (-m{)}^2-(4\times 1\times 9)=0\Rightarrow m^2-36=0\Rightarrow m^2=36\Rightarrow m=\pm \sqrt{36}\Rightarrow m=\pm 6$

(ii) If the roots are distinct then $b^2-4ac>0$, therefore,

$b^2-4ac>0\Rightarrow (-m{)}^2-(4\times 1\times 9)>0\Rightarrow m^2-36>0\Rightarrow m^2>36\Rightarrow m>\pm \sqrt{36}\Rightarrow m>\pm 6$

(iii) If the roots are imaginary then $b^2-4ac<0$, therefore,

$b^2-4ac<0\Rightarrow (-m{)}^2-(4\times 1\times 9)<0\Rightarrow m^2-36<0\Rightarrow m^2<36\Rightarrow m<\pm \sqrt{36}\Rightarrow m<\pm 6$

Hence $m=\pm 6$ if the roots are equal, $m>\pm 6$ if the roots are distinct and $m<\pm 6$ if the roots are imaginary.