Question

Solve the following equations:
$y+\sqrt{x^2-1}=2$

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

• A. $x=2;y=5$
• B. $x=1;y=2$
• C. $x=\frac{5}{3};\frac{2}{3}$
• D. $x=3;y=3$

Answer 1

Answer: (B), (C)

The correct options are
B $x=1;y=2$
C $x=\frac{5}{3};\frac{2}{3}$

$⟹$ Let $x+1=p,x-1=q$

$\therefore x^2-1=pq,p+q=2x$

$y+\sqrt{pq}=2⟶\left(I\right)$

$\sqrt{p}-\sqrt{q}=\sqrt{y}⟶\left(II\right)$
Squaring on both sides on equation $\left(II\right):-$

$p+q-2\sqrt{pq}=y⟶\left(III\right)$

$2y+2\sqrt{pq}=2\times 2⟶\left(IV\right)$ [multiplying equation (I) by 2]

$2y+p+q=y=4$$2x=4-y⟶\left(V\right)$

$y-2=-\sqrt{x^2-1}$ from equation $\left(I\right)$

So, solving :-
${(y-2)}^2=x^2-1$

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

$y^2+5-4y=x^2⟶\left(VI\right)$

$y^2+5-4y={\left(\frac{4-y}{2}\right)}^2$

$4y^2+20-16y=y^2+16-8y$

$3y^2-8y+4=0$

$3y^2-6y-2y+4=0$

$(3y-2)(y-2)=0$

$\therefore x=1,y=2$

$x=\frac{5}{3},y=\frac{2}{3}$