Question

If the equation $x^2+ax+b=0$ has distinct real roots and $x^2+a|x|+b=0$ has only one real root, then which of the following is true?

• A. $b=0,a>0$
• B. $b=0,a<0$
• C. $b>0,a<0$
• D. $b<0,a>0$

Answer 1

The correct option is A $b=0,a>0$

$x^2+ax+b=0$ has distinct real roots.
Hence
$a^2-4b\ge 0$ ...(i)
$a^2\ge 4b$
$aϵ(-\infty ,-2\sqrt{b}]\cup [2\sqrt{b},\infty )$
And
$|x{|}^2+a|x|+b=0$ has only one real root.
Then
$(|x|+\frac{a}{2}{)}^2-\frac{a^2}{4}+\frac{4b}{4}=0$
$(|x|+\frac{a}{2}{)}^2=\frac{a^2-4b}{4}$
$|x|=\frac{a\pm \sqrt{a^2-4b}}{2}$
Now
$\sqrt{a^2-4b}<a$ (considering b as positive)
Yet we have only one real root.
Now a cannot be negative in any case, as negative value of a wont give any root in second case.
Hence $b=0$ and $a>0$
$|x|$ can only be positive,
Hence
$x=\frac{a+\sqrt{a^2-4b}}{2}$
$x=\frac{a+a}{2}$
$x=a$