# nature of roots of Quadratic equation

If the roots of the equation $$ax^2 + 2bx + c=0$$ are real and disrinct, then show that the roots of the equation $$x^2-2(a+b)x+a^2+b^2+2c^2=0$$ are non-real complex numbers.

For the first equation discriminant >0, which gives $$b^2-ac>0$$.

for the second equation we have to show that discriminant <0. that is where I am facing problem.

$$\Delta = 4(a+b)^2-4(a^2+b^2+2c^2)= 8(ab-c^2)$$

• Compute the discriminant. – Wuestenfux May 31 at 7:29

I think the second equation it's $$x^2-2(a+c)x+a^2+c^2+2b^2=0.$$ If so, your reasoning gives a proof.
You can not show what you have to show, since, if $$a=b=1$$ and $$c=0$$, then both equations have real and distinct roots.