Question

Find the value of $x$ satisfying the equation $|||x^2-x+4|-2|-3|=x^2+x-12$

Answer 1

$|||x^2-x+4|-2|-3|=x^2+x-12$

Now consider

$f\left(x\right)=x^2-x+4$

Discriminant is $1-16=-15<0$ . Hence $f\left(x\right)=x^2-x+4>0$ for $\forall xϵR$

$||x^2-x+4-2|-3|=x^2+x-12\Rightarrow ||x^2-x+2|-3|=x^2+x-12$

Again consider ,

$f\left(x\right)=x^2-x+2$

Discriminant is $1-8=-7<0.$ Hence $f\left(x\right)=x^2-x+2>0\forall xϵR$

$\therefore |x^2-x+2-3|=x^2+x-12\Rightarrow |x^2-x-1|=x^2+x-12$

Again consider,

$f\left(x\right)=x^2-x-1$

Discriminant is $1+4=5>0$

So the roots are $x=\frac{1\pm \sqrt{5}}{2}$

$\therefore$ Case$1xϵ(\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2})$

$\Rightarrow -x^2+x+1=x^2+x-12\Rightarrow 13=2x^2\Rightarrow x=\sqrt{\frac{13}{2}}=\sqrt{6.5}\approx 2.54$

But $xϵ(\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2})$ i.e., $xϵ(-0.618,1.618)$

$\therefore x\cong 2.54$ is not admissible.

Case$2$ x\quad \epsilon \quad (-\infty ,\cfrac { 1-\sqrt { 5 } }{ 2 } ]\cup [\cfrac { 1+\sqrt { 5 } }{ 2 } ,\infty )$

$x^2-x-1=x^2+x-12\Rightarrow 2x=11x=5.5$