Question

If the roots of the equations ax²+2bx+c=0and bx²-2vacx+b=0are simultaneously real, then prove that b2 = ac.

Answer 1

Roots of the given equations ax^2 + 2bx +c =0 and bx^2 - 2√ac x +b =0 are simultaneously real.

1. We know from 1st equation for it to be real,

4b^2- 4ac >= 0 and for equation 2

(-2√ac)^2 - 4b^2 >= 0.

2. As the roots are simultaneously real we can write 4b^2- 4ac = (-2√ac)^2 - 4b^2

3. By solving we get 8ac= 8b^2

i.e b^2 = ac.

Hence proved.