Friday, May 29, 2015

Solve $\sqrt{x+16}+\sqrt[4]{x+16}=12$.

Solve $\sqrt{x+16}+\sqrt[4]{x+16}=12$.

For some students, they would think to square the given equation three times to get rid of the fourth root and square root terms:










$2304(25600+320x+x^2)(x+16)=x^4+1790 x^3+870625 x^2+62292000 x+1211040000$

$2304 x^3+774144 x^2+70778880 x+943718400=x^4+1790 x^3+870625 x^2+62292000 x+1211040000$

$x^4-514 x^3+96481 x^2-8486880 x+267321600 = 0$

If you're motivated enough and determined to solve it and believe that you have done no wrong or used the more complicated method since you hold fast to the principle that in solving math challenge, solving it is what matters, then you might realize the two integer real roots that satisfy the equation (*) above are $65$ and $240$ and if you are not aware that the function represented by the LHS of the given equation is strictly increasing, you would perform the long division to look for its other roots:

$x^4-514 x^3+96481 x^2-8486880 x+267321600 = 0$

$(x-65) (x-240) (x^2-209 x+17136)=0$

$(x-65) (x-240)\left(\left(x-\dfrac{209}{2}\right)^2+\dfrac{24863}{4}\right)=0$

We must check and verify if the answers $x=65,\,240$ are the solution.

For $x=65$:




For $x=240$:



 $16+4\ne 12$

Therefore, the only solution for this problem is $x=65$.

But, do you think we need to square the given equation for that many times to solve for this problem? No, this problem can be tackled elegantly if one uses the substitution skill.

If one let $u=\sqrt[4]{x+16}$, that implies $u^2=\sqrt{x+16}$, also $u\ge 0$ and the given equation becomes



Since $u\ge 0$, we have only one solution for $u$, and that is $u=3$.

Back substitute $u=3$ into $u^2=\sqrt{x+16}$, we have:



$x=65$ and we're done.

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