Question

If $f\left(g\right(a\left)\right)=0$ where $g\left(x\right)=\frac{x}{4}+2$ and $f\left(x\right)=|x^2-3|$, find the possible value of $a.$

• A. $-8+4\sqrt{3}$
• B. $-(8+4\sqrt{3})$
• C. $6$
• D. $18$

Answer 1

Answer: (A), (B)

$-8+4\sqrt{3}$
$-(8+4\sqrt{3})$

Given $g\left(a\right)=\frac{a}{4}+2,f\left(x\right)=|x^2|-3$ and $f\left(g\right(a\left)\right)=0$

Now, $f\left(g\right(a\left)\right)=|g(a{)}^2-3|=|(\frac{a}{4}+2{)}^2-3|$

$=|\frac{a^2}{16}+a+4-3|=|\frac{a^2}{16}+a+1|=\frac{|a^2+16a+16|}{16}$

but$f\left(g\right(a\left)\right)=0$

$\Rightarrow \frac{|a^2+16a+16|}{16}=0$

$\Rightarrow a^2+16a+16=0$ $[\because |x|=0\Rightarrow x=0]$

$\Rightarrow a=\frac{-16\pm \sqrt{{16}^2-4\left(1\right)\left(64\right)}}{2}$

$\Rightarrow a=\frac{-16\pm \sqrt{256-64}}{2}=\frac{-16\pm 8\sqrt{3}}{2}$

Hence, $a=-(8+4\sqrt{3})$ and $a=-8+4\sqrt{3}$.