Question

If the roots of the equations `ax^2+2bx+c=0` and `bx^2-2sqrtacx+b=0`are simultaneously real then prove that `b^2=ac`

Answer 1

It is given that the roots of the equation `ax^2+2bx+c=0` are real 

∴` D_1=(2b)^-4xxaxxc=0` are real. 

⇒`4(b^2-ac)≥0` 

⇒` -4(b^2-ac)≥0` 

⇒ b^2-ac≥0           ................(1) 

Also, the roots of the equation `bx^2-2sqrtacx+b=0` are real. 

∴` D_2=(-2sqrtac)^2-4xxbxxb≥0` 

⇒`4(ac-b^2)≥0` 

⇒`-4(b^2-ac)≥0`

⇒`b^2-ac≥0`          ....................(2) 

The roots of the given equations are simultaneously real if (1) and (2) holds true together.
This is possible if 

`b^2-ac=0` 

⇒`b^2=ac`