Question

The number of real roots of the equation $x^{4}+\sqrt{x^{4}+20}=22$ is

• A. 4
• B. 2
• C. 0
• D. 1

Answer 1

Answer: (A)

$\quad\quad\quad\quad x^{4}+\sqrt{x^{4}+20}=22$
or$\quad \quad \quad x^{4}-22=-\sqrt{x^{4}+20}$
Put $x^{4} = y$ and square both the sides.
$\quad \quad \quad \quad \left(y-22\right)^{2}=y+20$
$\quad \quad \quad \quad y^{2}+484-44y=y+20$
$\quad \quad \quad \quad y^{2}+45y+464=0$
$\quad \quad \quad \quad y^{2}-29y-16y+464=0$
$\quad \quad \quad \quad \left(y-29\right)\left(y-16\right)=0$
$\quad \quad \quad \quad y=16,29$
$\quad \quad \quad \quad\therefore x^{4}=16.29$ or x$=\pm2,\pm2.31$