Prove, that two equations are equivalent

EDIT: Missed something very important! Sorry! We have $x^4+1=2(2x-1)^{1/4}$ not $x^4+1=2\sqrt{2x-1}$. One friend of mine told me that the equation $x^4+1=2(2x-1)^{1/4}$, where $x\geq \frac{1}{2}$ is equivalent to $$x^4+1=2x$$. How did he obtain this? The equation $x^4+1=2(2x-1)^{1/4}$ is equivalent to $\frac{x^4+1}{2}=\sqrt{2x-1}$ and we can consider two functions: One is $f:[\frac{1}{2},\infty)\rightarrow[0,\infty)$, $f(x)=(2x-1)^{1/4}$ and $g:[0,\infty)\rightarrow [\frac{1}{2},\infty), g(x)=\frac{x^4+1}{2}$. $g$is the inverse of$f$ so they can only intersect on the line $y=x$. Then we can consider the equation $x^4+1=2x$ Is that enough?

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What you can say is that $$x^4 +1 = 2\sqrt{2x - 1}\tag{1}$$ $$x^4 + 1 = 2x\tag{2}$$

share one solution: when $x = 1$, both equations are satisfied.

But each equation has a second solution not shared by the other.

• Real solutions to $(1)$ are $x = 1,\;x\approx 0.68682$.

• Real solutions to $(2)$ are $\;x = 1,\;\text{ and }\;x \approx 0.54369.$

Hence, equations $(1)$ and $(2)$ cannot be equivalent.

Note also that $$2x = 2\sqrt{2x-1} \iff x = 1.$$ The equality holds if and only if $x = 1$, and nowhere else.

We can see each of $f(x) = x^4 + 1, \;g(x) = 2\sqrt{2x - 1},\;h(x) = 2x$ in the following graph (refer to top graph only):

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Edited the question...sorry! Give it another try! –  Karl Z. Mar 1 at 16:13
Re: your edit. Actually, the two solutions to $x^4 + 1=2(2x - 1)^{1/4}$ are precisely the same two solutions to $x^4 + 1 = 2x$, which are precisely the same solutions to $2x = 2(2x-1)^{1/4}$. So, while the equations are not strictly equal, they are equivalent in the sense that they determine identical solutions. –  amWhy Mar 1 at 16:24
Yes, I know, but how to prove that they determine identical solutions? –  Karl Z. Mar 1 at 16:30
Have you tried graphing them? $$(2x)^4 = 2^4(2x-1) \iff 16x^4 = 32x - 16 \iff 16x^4 - 32 x + 16 = 0\iff x^4 - 2x + 1 = 0$$ Try finding the solutions to this. –  amWhy Mar 1 at 16:34
No, I need an algebric method... –  Karl Z. Mar 1 at 16:35

This is false. Suppose the equivalence as given: $x^4 + 1 = 2\sqrt{2x-1}$ exactly when $x^4 + 1 = 2x$ for $x \ge 1/2$. In particular, this means that if we assume $x^4 + 1 = 2x$, then $$x^4 + 1 = 2\sqrt{2x-1} = 2\sqrt{(x^4 +1)-1} = 2x^2,$$ so $x^4 - 2x^2 + 1 = (x^2 -1)^2 = 0$ and $x = 1$.

But noting the polynomial $p(x) = x^4 - 2x + 1$ has $p(1) = 0$ and dividing by the factor $x - 1$ gives a quotient $q(x) = x^3 + x^2 + x - 1$. Then $q(1/2) < 0$ and $q(1) > 0$ means $q$ has a root between 1/2 and 1, and so therefore does $p$. In other words, $x^4 + 1 = 2x$ and $x \ge 1/2$ does not imply $x = 1$, as we found by taking your friend's assumption.

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