Given a quadratic equation, find values of $a$ such that the equation will have four different solutions.

Question

I was given this problem:

For what values of a, equation $2a(x + 1)^2 − | x + 1| + 1 = 0 $ has four different solutions?

Can someone tell me a way to start this problem. (I was thinking of discriminant, but it didn’t work)

Answer 1

Hint

Assuming $x$ is real $$(x+1)^2=|x+1|^2$$

Again each non zero value of $y=|x+1|$ will corresponds to two distinct real values of $x$

So, how many distinct roots we need from $$2ay^2-y+2=0$$

Answer 2

Hint:

Split it into $x\ge-1$ and $x\lt-1$. Use the quadratic formula.

Answer 3

Let $t=|x+1|$. Then $2a(x+1)^2-|x+1|+1=0\implies 2at^2-t+1=0$.

$t$ naturally has two real solutions, so if $2at^2-t+1=0$ has two distinct real solutions, then the original equation will have four distinct real solutions.

For $2at^2-t+1=0$, we have \begin{align} \Delta=(-1)^2-4(2a)(1)&>0\\ 1-8a&>0\\ a&<\frac18 \end{align}

But, we also need to consider that if $a$ is negative, then the vertical reflection will cancel out the effect of the absolute value.

Therefore, we want $\boxed{0<a<\dfrac18}$. What part of your solution using the discriminant did not work?

Here is a visual.

Answer 4

$$ 2au^2 = |u|-1\Rightarrow 2a|u|^2-|u|+1=0 $$

then

$$ |u|=\frac{1\pm \sqrt{1-8a}}{2a} $$

so we choose $0 < a\lt \frac 18$