Question

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

• A. $4$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is A $4$

$x^4+\sqrt{x^4+20}=22$

$\sqrt{x^4+20}=22-x^4$

$x^4+20=(22-x^4{)}^2$

$x^4+20=484-44x^4+x^8$

$x^8-45x^4+464=0$

$x^8-29x^4-16x^4+464=0$

$(x^4-29)(x^4-16)=0$

$x^4=29$ or $x^4=16$

$x^2=\pm \sqrt{29}$ $x^4=\pm \sqrt{16}$

$x^2=\sqrt{29}$

$\downarrow$

$x=\pm \sqrt[4]{429}$

$x^2=-\sqrt{29}$

$\downarrow$

No real solutions

$x^2=4$

$\downarrow$

$x=\pm \sqrt{4}$

$=\pm 2$

$x^2=-4$

$\downarrow$

No real solutions

$\therefore$ The real solutions are $\pm 2$ and $\pm \sqrt[4]{429}$

Hence there are 4 real solutions.