Question

The number of real roots of $(x+3{)}^4+(x+5{)}^4=16$ is

• A. $0$
• B. $2$
• C. $4$
• D. None of these

Answer 1

The correct option is B $2$

Substitute:

$y=x+4$ ____ (1)

Then we have:

$(y-1{)}^4+(y+1{)}^4=16$

Expanding binomials, odd powers of yy cancel, giving

$2y^4+12y^2+2=16$

$\Rightarrow 2y^4+12y^2+2=16$

solve this quadratic in y2y2 for y2y2 then take square roots on both sides:

$y=\pm \sqrt{\frac{-12\pm \sqrt{{12}^2+4\times 2\times 14}}{4}}$

substitute into (1)(1):

$x=\pm \sqrt{\frac{-12\pm \sqrt{{12}^2+4\times 2\times 14}}{4}}$

If you evaluate this for all four ±± combinations the two real roots turn out to be −5−5 and −3−.