Real Solutions of the expression $\sqrt[2]{x+3-4\sqrt[2]{x-1}}+\sqrt[2]{x+8-6\sqrt[2]{x-1}}-1=0$

Question

I recently stumbled across the following question:

Find all the real solutions $x$ of the expression $\sqrt[2]{x+3-4\sqrt[2]{x-1}}+\sqrt[2]{x+8-6\sqrt[2]{x-1}}-1=0$

It was given by one of my teachers (he was teaching us a few problems involving real and rational numbers).

Although I used a few typical algebraic manipulations (such as shifting terms from LHS to RHS and squaring both sides) all of them repeatedly led me towards wrong results. When I plugged the expression $\sqrt[2]{x+3-4\sqrt[2]{x-1}}+\sqrt[2]{x+8-6\sqrt[2]{x-1}}-1$ into Desmos, I came to see that all real numbers from 5 to 10 gave the value of zero when plugged into the expression. Could anyone show (through algebraic manipulations) why this is the case?

Answer 1

$\sqrt[2]{x+3-4\sqrt[2]{x-1}}+\sqrt[2]{x+8-6\sqrt[2]{x-1}}=1$ Let $x=1+t^2$, then we re-write the eq. as $$\sqrt{(t-2)^2}+\sqrt{(t-3)^2}=1$$ $$\implies |t-2|+|t-3|=1$$ When $2\le t \le 3$. then the equation becomes $$ (t-2)+(3-t)=1\implies 1=1$$ So all roots are $2\le t\le 3$ \implies $5 \le x \le 10.$ Hence this equation has all numbers in $[5,10]$ as infinitely many roots.

Similarly one can check no roots in $(-\infty,5)$ and no root in $(10,\infty)$.

Answer 2

Looks like a typical high school problem easy one.

Hint: Use $a^2\;+\;b^2\;-2ab\;=\;(a-b)^2$

Solution: $\sqrt{(\sqrt{x-1})^2\;+(2)^2-2(2)(\sqrt{x-1})}\;+\;\sqrt{(\sqrt{x-1})^2\;+(3)^2\;-2(3)(\sqrt{x-1})}\;-1\;=\;0$

$\sqrt{x-1}\;-\;2\;+\;\sqrt{x-1}\;-\;3\;-1\;=\;0$

$2\sqrt{x-1}\;=\;6$

$x\;=\;10$

It is quite clear that any use of modulus or checking radicand to be non negative is futile.

Hence it has a real solution