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

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

Student 4
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    It is probably easiest to consider the cases $x \geq 0$ and $x <0$ separately. In the latter case we have $f(x) = x^2 -(a+3)x+4=0$. – Slugger Jun 28 '21 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 – Student 4 Jun 28 '21 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 '21 at 21:10

1 Answers1

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

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 $a \in (-\infty , -7)$

Math Lover
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  • I don't understand why $\lvert x\rvert$ matter if $a> -3$Could you please explain this? – Student 4 Jun 28 '21 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. – Math Lover Jun 28 '21 at 18:04
  • Oh.... Thank you for the solution@Math lover.If I may ask what topics do you have expertise in? – Student 4 Jun 28 '21 at 18:09