# Find real x such that $\sqrt[4]{97-x}+\sqrt[4]{x}=5$

I tried this: $$97-x=(5-\sqrt[4]{x})^4$$ But the expression became too complicated to manipulate, so maybe there's a better way to do this? I believe it can be done using Vieta's relations because the whole Exercise in the textbook was about roots of equations.

• Like math.stackexchange.com/questions/616536/… if $$\sqrt[4]x= a,\sqrt[4]{97-x}=b$$ $$ab$$ are the roots of $$(t-44)(t-6)=0$$ – lab bhattacharjee Apr 8 at 15:07
• Let $x=y^4$ and expand the binomial term. Then, you can use Cardano's formula to find the roots. – Mohammad Zuhair Khan Apr 8 at 15:09
• Thanks, why didn't I think of that! – Sujal Motagi Apr 8 at 15:13
• @labbhattacharjee Why are the roots of the equation the roots $ab$. Can you please explain? – lkfalkfda Apr 8 at 15:24
• @Okbko, Please follow my post in the link. Let me if you have any further query – lab bhattacharjee Apr 8 at 15:25

Let $$t=\sqrt[4]x$$. Then, the equation can be written as,

$$(5-t)^4-(97-t^4)=0$$

Factorize to get

$$(t-2)(t-3)(t^2-5t+44)=0$$

which leads to $$t=2,\>3$$. Thus, the solutions are $$x=t^4=16,\>81$$.

• Oh, this is elegant. – Sujal Motagi Apr 8 at 16:14

Let $$\sqrt[4]{97-x}=a$$ and $$\sqrt[4]x=b$$.

Thus, $$a+b=5$$ and $$a^4+b^4=97.$$ But $$a^4+b^4=(a^2+b^2)^2-2a^2b^2=((a+b)^2-2ab)^2-2a^2b^2=$$ $$=(25-2ab)^2-2a^2b^2=625-100ab+2a^2b^2.$$ Thus, $$a^2b^2-50ab+264=0$$ or $$(ab-25)^2=361,$$ which gives $$ab=44,$$ which is impossible, or $$ab=6.$$ Can you end it now?

I got the following answer. $$\{16,81\}$$

An idea, starting from your equation: $$97-x=(5-\sqrt[4]{x})^4$$ is to develop the right side of the equation as follows: $$(a-b)^{4}=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$$ and then we set $$a=\sqrt[4]{x}$$ and we can rewrite the whole equation as a fourth order polynomial: $$97-a^4=625-500a+150a^2-20a^3+a^4$$ which can be simplified into: $$-2a^4 + 20a^3 -150a^2+500a-582 =0$$

Hint : Since the expression on the LHS is a increasing for the $$0 and decreasing for $$48.5 .Therefore , checking the values of the expression at $$x=0$$,$$x=48.5$$,$$x=97$$ , it is clear that this equation has only two real solution . Further , with a little inspection it is clear that if $$k$$ is a solution then $$97-k$$ is also a solution. Inspecting further(by error and trial with intentions of getting integer terms) $$x=16$$ is a solution , and thus $$x=81$$ is also a solution . And these are the only two real solution from the above argument.

• How can I prove that 16 and 81 are the only real soutions? – Sujal Motagi Apr 8 at 15:38
• Never mind, I got it. – Sujal Motagi Apr 8 at 15:47

By cheating:

It is likely that the solution is an integer. We will try all fourth powers until the expression under a radical is also a fourth power.

$$0\to97,\\1\to96,\\16\to81.$$

Bingo, $$3+2=5$$ !

By setting $$x=t^4$$, you have

$$\sqrt{97-t^4}+t=5$$

which implies

$$97-t^4=(t-5)^4.$$

You will get a quartic polynomial with extreme coefficients $$2$$ and $$5^4-97=528=2^4\cdot3\cdot11.$$ and you can use the rational root theorem, but there are many options, though you need only try those below $$\sqrt[4]{97}$$.

• That's what I did later :). And I think you mean 1-->96 not 76. – Sujal Motagi Apr 8 at 15:37

The right hand side has derivative $$-\frac{(5-x^{1/4})^3}{x^{3/4}},$$ which is negative for all $$x$$ satisfying $$0 Thus the right hand side is decreasing in this range. The second derivative is given by $$\frac{3(5-x^{1/4})^2}{4x^{3/2}}+\frac{3(5-x^{1/4})^3}{4x^{1/4}},$$ which is also positive in the range considered. Thus it follows that the right hand side of the equation is convex. Thus there are at most two intersections of the curve $$y=\text{RHS}$$ with the line $$y=97-x.$$ This proves that there are at most two real solutions of the equation.

Indeed, some of the previous answers have found two solutions. This completes the case.

• I'm 13 years old, so calculus is not really my best subject :) – Sujal Motagi Apr 8 at 16:03
• @SujalMotagi Oh, I didn't know this. – Allawonder Apr 8 at 16:39