On the equation $x^2+|x|-6=0$

Question

Which of the following are true given the following equation

$$|x|^2+|x|-6=0$$

$1$. It has $4$ roots

$2$. The sum of the roots is $-1$

$3$. The product of the roots is $-4$

$4$. The product of the roots is $-6$

Only one of the options is correct.

Answer 1

Assuming $x\geq 0$, the equation will be $x^2+x-6=0$ which has two solutions $\frac{-1\pm \sqrt{25}}{2}$ which only $2$ is acceptable.

Assuming $x<0$, the equation will be $x^2-x+6=0$ which has two solutions $\frac{1\pm \sqrt{25}}{2}$ which only $-2$ is acceptable.

So, among your options, it seems that only the $\color{red}{third}$ one is true.

Answer 2

Hint: look at $x\geq 0: x^2+x-6=0$ (use quadratic formula and see how many solutions are positive). and $x <0: x^2-x-6=0$(use quadratic formula and see how many solutions are negative)

Answer 3

It has $2$ roots, since setting $u=\lvert x\rvert$, it amounts to solving non-negative roots of $u^2+u-6$, which has roots $ 2$ and $-3$. Hence the roots are $\pm 2$, and their product is $-4$.

Answer 4

Let $|x| = t \ge 0$

$t^2+t-6=0$

$t=\frac{-1 \pm\sqrt{1+24}}{2}$

$t=\frac{-1 \pm5}{2}$

$t=\frac{4}{2} || \frac{-6}{2}$

$t=2||-3$

But we know that $t \ge 0$

Therefore, $t=2$

After that, $x=\pm 2$

Thus, the answer is option $3$.

The product of the roots is −4

Answer 5

First of all, you can rewrite the equation as $x^2 + |x|- 6 = 0$

$\text{CASE}\; 1: x > 0$

\begin{align} x^2 + x - 6 &= 0 \\ (x + 3)(x - 2) &= 0 \\ x &= 2 \end{align}

$\text{CASE}\; 2: x < 0$

\begin{align} x^2 - x - 6 &= 0 \\ (x - 3)(x + 2) &= 0 \\ x &= -2 \end{align}

So the product of the roots is $-4$.

The changed question.

\begin{align} |x|^2+|x|-6 = 0 \\ (|x|+3)(|x|-2) &= 0 \\ |x|&=2 \\ x &\in \{2, -2\} \\ \end{align}

Hence the product of the roots is $-4$.