Question

If $\sqrt{x-1}-\sqrt{x+1}+1=0$, then 4x equals

• A. $5$
• B. $4\sqrt{-1}$
• C. $0$
• D. $1\frac{1}{4}$
• E. no real value

Answer 1

The correct option is A $5$

$\sqrt{x-1}-\sqrt{x+1}+1=0$

$\sqrt{x-1}-\sqrt{x+1}=-1$

take square both side

$[\sqrt{(x-1)}-\sqrt{(x+1)}{]}^2=(-1{)}^2$

apply $(a-b{)}^2=a^2+b^2-2ab$

$(x-1)+(x+1)-2\sqrt{(x-1)(x+1)}=1$

$2x-1=2\sqrt{x^2-1}$

again take square both side

$(2x-1{)}^2=(2\sqrt{x^2-1}{)}^2$

$4x^2+1-4x=4(x^2-1)$

$4x^2+1-4x=4x^2-4$

$1+4=4x$

$4x=5$