# Find the range of $a$ for the following quadratic equation.

$$f(x) = x^2 +(a+3)\lvert x \rvert + 4 = 0$$

This is the quadratic equation They have given condition that find the range of $$a$$ for which the roots are real

So what I did was

$$D\geq 0$$

Solved the condition and got

$$(a+7)(a-1) \geq 0$$

So $$a \in (-\infty , -7] \cup [1,\infty)$$

But the given key is only $$a \in (-\infty , -7]$$

Could someone explain where I went wrong so $$[1,\infty)$$ does not come in the solution

• It is probably easiest to consider the cases $x \geq 0$ and $x <0$ separately. In the latter case we have $f(x) = x^2 -(a+3)x+4=0$. Jun 28, 2021 at 16:55
• Anyway that does not matter as we put $b^2 = (a+3)^2 =[-(a+3)]^2$ they will give same value irrespective of x being negative Jun 28, 2021 at 17:04
• @Student4 $D \ge 0$ is the condition for $x^2+(a+3)x+4=0$ to have real roots. But your equation is $y^2+(a+3)y+4=0$ where $y=|x| \ge 0$. For this one to have real roots, the quadratic needs to have positive real roots. Positiveness is the additional condition that your solution missed.
– dxiv
Jun 28, 2021 at 21:10

$$f(x) = x^2 +(a+3)\lvert x \rvert + 4 = 0$$

If $$\alpha$$ and $$\beta$$ are roots of the equation,

$$\alpha + \beta = - (a+3)$$

$$\alpha \beta = 4$$

If $$a \gt - 3,$$ the quadratic has both negative roots but $$|x|$$ cannot be negative.

Hence the only solution that works is $$a \in (-\infty , -7)$$

• I don't understand why $\lvert x\rvert$ matter if $a> -3$Could you please explain this? Jun 28, 2021 at 17:58
• When $a \in (1, \infty)$, $a \gt -3$ so you only have negative roots. Take example when $a = 1$. We get $x^2 + 4 |x| + 4 = 0 \implies (|x| +2)^2 = 0$. That gives $|x| = - 2$ which is not possible. Jun 28, 2021 at 18:04
• Oh.... Thank you for the solution@Math lover.If I may ask what topics do you have expertise in? Jun 28, 2021 at 18:09