Quadratic Problem - 16th Dec


!!ax^2 + arx + ar^2 = 0!!

If !!a!! and !!r!! are real.

Then roots of the equation are?

!!a)" "!!Always imaginary

!!b)" "!!Both real roots

!!c)" "!!Only one real root

!!d)" "!!Depends on !!r!!


Given equation:!!ax^2 + arx + ar^2 = 0!!

To check for nature of roots, we need to find the value of discriminant !!(D)!!.

!!D = (a^2)(r^2) - 4(a)(ar^2) = a^2(r^2 - 4r^2) = -3(a^2)(r^2)!!

Hence !!D!! will always remain negative.

This proves roots of the equation are always imaginary.