Question

Find the discriminant for the quadratic equation represented by $\left(\frac{m+n}{2}\right)x^2+(m+n)x+\left(\frac{m+n}{2}\right)=0$ and determine the nature of roots. $(m,n>0)$

• A. ${\text{D = (m+n)}}^2;\text{Roots are real and equal}$
• B. ${\text{D = (m-n)}}^2;\text{Roots are non - real}.$
• C. $\text{D = mn},\text{Roots are real and distinct.}$
• D. $\text{D =0};\text{Roots are real and equal}$

Answer 1

The correct option is D $\text{D =0};\text{Roots are real and equal}$

$\left(\frac{m+n}{2}\right)x^2+(m+n)x+\left(\frac{m+n}{2}\right)=0$

$a=\left(\frac{m+n}{2}\right),b=(m+n),c=\left(\frac{m+n}{2}\right)$

$D=b^2-4ac$

$=(m+n{)}^2-4\times {\left(\frac{m+n}{2}\right)}^2$

$=(m^2+n^2+2mn)-4\left(\frac{m^2+n^2+2mn}{4}\right)$

$=(m^2+n^2+2mn)-(m^2+n^2+2mn)=0$

$\text{Since D=0},$

$\text{Hence, the roots will be real and equal}$

$\text{(d) is correct}$