Question

For what positive values of '$m$' roots of given equation is equal , distinct, imaginary $r^2-(m+1)r+4=0$

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 $r^2-(m+1)r+4=0$ is in the form $a{x}^2+bx+c=0$ where $a=1,b=-(m+1)$ and $c=4$.

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

$b^2-4ac=0\Rightarrow (-(m+1){)}^2-(4\times 1\times 4)=0\Rightarrow (m+1{)}^2-16=0\Rightarrow (m+1{)}^2=16\Rightarrow m+1=\pm \sqrt{16}\Rightarrow m+1=\pm 4\Rightarrow m+1=4,m+1=-4\Rightarrow m=4-1,m=-4-1\Rightarrow m=3,m=-5$

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

$b^2-4ac>0\Rightarrow (-(m+1){)}^2-(4\times 1\times 4)>0\Rightarrow (m+1{)}^2-16>0\Rightarrow (m+1{)}^2>16\Rightarrow m+1>\pm \sqrt{16}\Rightarrow m+1>\pm 4\Rightarrow m+1>4,m+1>-4\Rightarrow m>4-1,m>-4-1\Rightarrow m>3,m>-5$

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

$b^2-4ac<0\Rightarrow (-(m+1){)}^2-(4\times 1\times 4)<0\Rightarrow (m+1{)}^2-16<0\Rightarrow (m+1{)}^2<16\Rightarrow m+1<\pm \sqrt{16}\Rightarrow m+1<\pm 4\Rightarrow m+1<4,m+1<-4\Rightarrow m<4-1,m<-4-1\Rightarrow m<3,m<-5$

Hence $m=3,m=-5$ if the roots are equal, $m>3,m>-5$ if the roots are distinct and $m<3,m<-5$ if the roots are imaginary.