Finding $a$ yielding minimum value for quadratic root expression $(x_1+2x_2)(x_2+2x_1)$ The problem is:

We have the expression $(x_1+2x_2)(x_2+2x_1)$, where $x_1$ and $x_2$ are the roots of $$f(x)=x^2+ax+a+\frac{1}{5}$$
  Find the value(s) of $a$ yielding the least possible value for this expression.

My solution is: for readability's sake, $x_1=t, x_2=z$, then 
$$(x_1+2x_2)(x_2+2x_1)=(t+2z)(z+2t)= .. =2(t+z)^2-tz$$
Using Vieta's formulas, we get
$$2(t+z)^2-tz=2\left(\frac{-B}{A}\right)^2-\frac{C}{A}=2a^2-a-0.2$$
From here, the minimum point is the apex of the parabola:
$$a_0=\frac{-B}{2A}=\frac{1}{4}$$
Yet the textbook's answer is $$-\frac{1}{4}$$
Is there a mistake in my calculations? 
 A: Hint
We consider $$A=(x_1+2x_2)(x_2+2x_1)=5x_1 x_2+2(x_1^2+x_2^2)=5x_1x_2+2\big((x_1+x_2)^2-2x_1x_2\Big)$$ $$A=2(x_1+x_2)^2+x_1x_2$$ But, from $x^2+ax+a+\frac{1}{5}=0$ $$x_1+x_2=-a$$ $$x_1x_2=a+\frac 15$$ So, $$A=2a^2+a+\frac 15$$ $$A'=4a+1=0$$
I am sure that you can take from here.
A: Given $x_1$ and $x_2$ are the roots of $\displaystyle f(x)=x^2+ax+a+\frac{1}{5} = 0$
So $\displaystyle x_{1}+x_{2} = -a$ and $\displaystyle x_{1}\cdot x_{2} = a+\frac{1}{5}$
Now $$\displaystyle (x_1+2x_2)(x_2+2x_1) = x_{1}\cdot x_{2}+2\left[x_{1}^2+x_{2}^2\right]+4x_{1}\cdot x_{2} = 5x_{1}x_{2}+2\left[\left(x_{1}+x_{2}\right)^2-2x_{1}\cdot x_{2}\right]$$
So Let $$\displaystyle f(a) = -5\left(a+\frac{1}{5}\right)+2\left[a^2-2\left(a+\frac{1}{5}\right)\right] = 2a^2+a+\frac{1}{5}$$ 
So $$\displaystyle f(a) = 2\left[a^2+\frac{a}{2}+\frac{1}{10}\right] = 2\left[a^2+\frac{a}{2}+\frac{1}{4^2}-\frac{1}{4^2}+\frac{1}{10}\right] = 2\left[\left(a+\frac{1}{4}\right)^2+\frac{3}{80}\right]$$
Thus, the minimum of the fuunction $f(a)$ occurs when $\displaystyle \left(a+\frac{1}{4}\right)^2 = 0\Rightarrow a= -\frac{1}{4}.$
