Question

If 2|x+1|2 -− |x+1| = 3, then x =​

Answer 1

Answer in attachment also.

$\huge\tt{\red{\underline{Given:}}}$

★$2|x+1|^{2}-|x+1|=3$

$\huge\tt{\red{\underline{To\:\:Find:}}}$

★$The\:value\:of\:x$

$\huge\tt{\red{\underline{Concept\:\:Used:}}}$

★We would factorise the given equation and then try to find the value of x .

$\huge\tt{\red{\underline{Answer:}}}$

We have,

$\implies 2|x+1|^{2}-|x+1|=3$

$\implies 2|x+1|^{2}-|x+1|-3=0$

$\implies 2|x+1|^{2}-3|x+1|+2|x+1|-3=0$

(Using middle term splitting for factorising )

$\implies |x+1|(2|x+1|-3)+1(2|x+1|-3) =0$

$\implies (|x+1|+1)(2|x+1|-3)=0$

$\:\:\:\:\:\:\:\:\:\:\: \: \:\:\:\: (1) \:\:\:\:\:\:\:\: \:\:\:\: \:\:\:\:\:\:\:\: (2) \:\:\:\:$

Taking 1:

$\implies (|x+1|+1)=0$

${\underline{\boxed{\red{.°. |x+1|=-1}}}}$

This is a not possible cases the value of mod is never negative .

Taking 2:

$\implies (2|x+1|-3)=0$

$\implies 2|x+1|=3$

$\implies |x+1|=\dfrac{3}{2}$

On squaring both sides,

$\implies (|x+1|)^{2}=(\dfrac{3}{2})^{2}$

$\implies (x+1)^{2}=\dfrac{9}{4}$

$\implies (x^{2}+1+2x)=\dfrac{9}{4}$

$\implies 4(x^{2}+1+2x)=9$

$\implies 4x^{2}+4+8x-9=0$

$\implies 4x^{2}+8x-5=0$

$\implies 4x^{2}+10x-2x-5=0$

$\implies 2x(2x+5) -1(2x+5) =0$

$\implies (2x-1) (2x+5) =0$

${\underline{\boxed{\purple{.°. x =\dfrac{1}{2}, \dfrac{-5}{2}}}}}$

$\huge\orange{\boxed{Answer:x=\dfrac{1}{2}, \dfrac{-5}{2}}}$