Question

If $a,b,c$ are distinct, positive in $H.P$, then quadratic equation ${ax}^2+2bx+c=0$ has

• A. Two non-real roots such that their sum is real
• B. Two distinct real roots
• C. Two non real conjugate roots
• D. Two equal real roots

Answer 1

The correct option is B Two distinct real roots

Given that $a,b,c$ are distinct and real roots in HP.

Thus we can say $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in AP.

$\therefore \frac{1}{b}=\frac{\frac{1}{a}+\frac{1}{c}}{2}=\frac{a+c}{2ac}$

Thus $b=\frac{2ac}{a+c}$ ....(i)

Now given quadratic equation is $a{x}^2+2bx+c=0$

We know discriminant $D=b^2-4ac$

$=(2b{)}^2-4\times a\times c$

$=4b^2-4ac$

$=4\left[\frac{4a^2{c}^2}{(a+c{)}^2}\right]-4ac$

$=\frac{16a^2{c}^2}{(a+c{)}^2}-4ac$

$=4ac(\frac{4ac}{(a+c{)}^2}-1)$

$=4ac\left[\frac{4ac-(a+c{)}^2}{(a+c{)}^2}\right]$

$=4ac\left[\frac{-a^2-c^2+2ac}{(a+c{)}^2}\right]$

$=-4ac\left[\frac{(a-c{)}^2}{(a+c{)}^2}\right]$

Given that $a,b,c$ are positive, but overall result is negative.

Thus $D<0$

We can say roots are imaginary.