Question

Prove that the roots of the equation $b{x}^2+(b-c)x+b(b-c-a)=0$ are real if those of $a{x}^2+2bx+b=0$ are imaginary and vice versa.

Answer 1

For roots to be real

${(b-c)}^2-4.b.(b-c-a)>0$

Or

$(b^2+c^2-2bc)-4(b^2-bc-ab)>0$

Solving above equation,

$-3b^2+2bc+c^2+4ab>0$

$(2bc+c^2)-3b^2+4ab>0$

Making square of$(b-c)$

${(b-c)}^2-4b(b-c-a)>0$

Solving above

${(b-c)}^2-4b^2+4bc+4ab>0{(b+c)}^2-4b^2+4ab>0{(b+c)}^2>4(b^2-4ab)$

Now as per given condition, for roots to be imaginary

$4b^2-4ab<04(b^2-ab)<04b(b-a)<0$

Or

$(a-b)>0$