Prove that the roots of $g(x)$ are also real

Question

Suppose there are two equations $$f(x)=x^2+bx+c=0$$ and $$g(x)=x^2+bx+c(x+a)(2x+b)=0$$ It is given that $f(x)$ has two real roots. Then prove that $g(x)$ also has two real roots when $a\in \mathbb{R}$.

My work:

$$g(x)=x^2(2c+1)+x(b+bc+2ac)+abc=0$$ $$\Delta g(x)=(b+bc+2ac)^2-4(2c+1)(abc)$$ $$=b^2+b^2c^2+4a^2c^2+2b^2c-4abc^2$$ Now we have to prove that $b^2+b^2c^2+4a^2c^2+2b^2c-4abc^2\ge0$ or $b^2c-2abc^2\ge0$

The only information we have is that $b^2-4c\ge0$ or $b^2c\ge4c^2$ or $$b^2c-2abc^2\ge 4c^2-2abc^2$$ I can't continue from here. Any help is greatly appreciated.

EDIT I just saw a counter example to this question. But let's suppose that $a\ge \frac {b}2$. Will it be true in that case$?$

Answer 1

This is not true or there are conditions missing on $a,b,c$.

Take $b=1$ and $c=-2$ then $f(x)=x^2+x-2=(x-1)(x+2)$ has two real roots.

While $g(x)=3x^2+(4a+1)x+2a$ has discriminant $\Delta=16a^2-16a+1$ which can be negative.

For instance for $a=\frac 12$.

Answer 2

$$b^2+b^2c^2+4a^2c^2+2b^2c-4abc^2$$ $$\ge b^2+b^2c^2+4(\frac{b}{2})^2c^2+2b^2c-4\frac{b}{2}bc^2$$ $$=b^2(1+2c)$$ So it is true only for $c>-\dfrac{1}{2}$.

EDIT If $c=-\dfrac{1}{2}$ then $g(x)=x(\dfrac{b}{2}-a)-\dfrac{ab}{2}=0$ has less than two roots. Thanks @zwim!