# sum of all non real roots of the equation in a bi-quadratic equation

Consider the equation $$8x^4-16x^3+16x^2-8x+a=0\;\left(a\in \mathbb{R}\right)\;,$$ Then the sum of

all non real roots of the equation can be

$$\bf{OPTIONS::}\;\; (a)\;\; 1\;\;\;\;\;\; (b)\;\; 2\;\;\;\;\;\; (c)\; \displaystyle \frac{1}{2}\;\; \;\;\;\; (d)\;\; None$$

$$\bf{My\; Try::}$$ Let $$f(x)=8x^4-16x^3+16x^2-8x+a\;,$$ Then $$f'(x)=32x^3-48x^2+32x-8$$

And $$\displaystyle f''(x) = 96x^2-96x+32 = 96\left[x^2-x+\frac{1}{3}\right]=96\left[\left(x-\frac{1}{2}\right)^2+\frac{1}{12}\right]>0\;\forall x \in \mathbb{R}$$

So Using $$\bf{LMVT\;,}$$ We get $$f'(x)=0$$ has at most $$1$$ real roots and

$$f(x)=0$$ has at most $$2$$ real roots

Now How can i solve it after that, Help me

Thanks

Now How can i solve it after that

Noting that $f'(1/2)=0$ is a key.

We have $$f'(x)=8(4x^3-6x^2+4x-1)=8(2x-1)\left(2\left(x-\frac 12\right)^2+\frac 12\right)$$ So, $f(x)$ is decreasing for $x\lt 1/2$ and is increasing for $x\gt 1/2$.

By the way, $$f\left(\frac 12+s\right)=8s^4+4s^2-\frac 32+a$$ and so $f(1/2-\alpha)=0$ is equivalent to $f(1/2+\alpha)=0$.

Also, by Vieta's formula, the sum of roots is $-(-16)/8=2$.

If $a\le 3/2$, then $f(1/2)\le 0$, and the answer is $$2-(1/2-\alpha)-(1/2+\alpha)=\color{red}{1}$$

If $a\gt 3/2$, then $f(1/2)\gt 0$, and the answer is $\color{red}{2}$.