Location of roots (quadratic Equation). Conditions on $a$

Question

Find set of values of $a$ If $($$x^2$+$x$)$^2$+ $a$$($$x^2$+$x$) + $4$ $=$ $0$ has all four real roots where two of them are equal.

Here’s how I tried solving it :

I assumed $($$x^2$+$x$$)$ = $t$

$\Rightarrow$ $t$ $\in$ $[$$\frac{-1}{4}$, $\infty$$]$

The given equation $\Rightarrow$ $f(t)$ $=$ $t^2$ + $at$ + $4$ $=$ $0$

This equation will have real roots in the interval $[$$\frac{-1}{4}$, $\infty$$]$

For four real roots, both the roots of $f(t)$ = $0$ should lie in the interval $[$$\frac{-1}{4}$, $\infty$$]$

For two equal roots, either $f(t)$ should be a perfect square, or $t$ $=$ $\frac{-1}{4}$ should be one of the roots of $f(t)$ $=$ $0$ (Both roots should still lie in $[$$\frac{-1}{4}$,$\infty$$)$

Now I imposed the conditions, and solved it

The set of values of $a$ that I am getting is $a$ $\in$ $($$-\infty$, $-4$$]$

But I checked on Desmos, the equation $($$x^2$+$x$)$^2$+ $a$$($$x^2$+$x$) + $4$ $=$ $0$ has all four real roots, where two are equal, for only one value of $a$ $\in$ {$-4$}.

When $a$ $\in$ ($-\infty$, $-4$), the equation $($$x^2$+$x$)$^2$+ $a$$($$x^2$+$x$) + $4$ $=$ $0$ has four real and distinct roots. We want at least two equal roots, that happens only when $a$ equals ($-4$).

I’m not sure what my mistake is. Please help me with it, thanks a lot.

Answer 1

The solutions of $t^2+at+4=0$ are $\frac{-a\pm\sqrt{a^2-16}}{2}$, hence $$x^2+x+\frac{a\mp\sqrt{a^2-16}}{2}=0.$$ Now the left side has to be a complete square, from where $$\frac{a\mp\sqrt{a^2-16}}{2}=\frac14.$$

That’s the case for $a=\pm65/4$, that is $a=65/4$ as we have $a>-1/4$.

Answer 2

I can write $f(t)=t^2+\alpha t+4=(t-\beta)\cdot (t-\gamma)$ with $\beta,\gamma\in R$. I can say that: $$\left\{\begin{matrix} \beta\cdot\gamma=4 \\-(\beta+\gamma)=\alpha \end{matrix}\right.$$

Because of two solutions have to be equal $t-\beta$ has to be a square: this occours only if $\beta=-\frac{1}{4}$; in fact $x^2+x+\frac{1}{4}=(x+\frac{1}{2})^2$.

Thanks to the second condition, I have $\beta\cdot\gamma=4$: in other words $-\frac{1}{4}\cdot\gamma=4$; $\gamma=-16$.

In conclusion, there is only one possible solution: $\alpha=-(\beta+\gamma)=-(-16-\frac{1}{4})=\frac{65}{4}$.