Question

If $a$ and $b$ are the roots of $\sqrt{x^2-16}-(x-4)=\sqrt{x^2-5x+4}$, then

• A. $a+b=9$
• B. $ab-(a+b)=0$
• C. $a=b$
• D. $2a=b$

Answer 1

Answer: (A)

$a+b=9$

$\sqrt{x^2-16}-(x-4)=\sqrt{x^2-5x+4}$
or $\sqrt{x+4}\sqrt{x-4}-\sqrt{x-4}\sqrt{x-4}=\sqrt{x-4}\sqrt{x-1}$
or $\sqrt{x+4}-\sqrt{x-4}=\sqrt{x-1}$
or $x+4+x-4-2\sqrt{x^2-16}=x-1$
or $2\sqrt{x^2-16}=x+1$
or $4x^2-64=x^2+2x+1$
or $3x^2-2x-65=0$
or $(x-5)(3x+1)=0$
i.e., $x=5$ or $-\frac{13}{3}$
$\therefore a=4,b=5$
$\therefore a+b=9$.