A possible solution to $\sqrt {5-x}=5-x^2$ (without taking square from both sides)

Begin by subtracting $x$ from both sides:

$\sqrt{5-x}-x=(5-x)-x^2$

And render the difference of squares factorization

$(\sqrt{5-x}-x)(\sqrt{5-x}+x)=(5-x)-x^2$

By comparison we must have

$(\sqrt{5-x}-x)(\sqrt{5-x}+x)=\sqrt{5-x}-x$

and we are led to two possibilities:

Possibility 1: if the common factor $\sqrt{5-x}-x$ is nonzero we must have

$\sqrt{5-x}+x=1$,

from which

$5-x^2+x=1, x^2-x-4=0, x=(1-\sqrt{17})/2$

where the sign on $\sqrt{17}$is fixed by requiring $x^2\le 5$ because $\sqrt{5-x}=5-x^2$ must be nonnegative.

Possibility 2: The common factor is zero, in which case we simply have

$\sqrt{5-x}=x=5-x^2, x^2+x-5=0,x=(-1+\sqrt{21})/2$

where again $x^2\le 5$ to make $\sqrt{5-x}=5-x^2$ nonnegative.

Thus the solution set is $\{(1-\sqrt{17})/2,(-1+\sqrt{21})/2\}$.


Hint: Write $t=\sqrt{5-x}\geq 0$ so $x=5-t^2$ and now $$t=5-(5-t^2)^2$$ so you have to solve $$ f(f(t))=t\;\;\;(*)$$

where $f(t) = 5-t^2$. Clearly the fixed points of $f$ satisfies the equation $(*)$, so solution to $t^2+t-5=0$ are two solution to $(*)$: $$t_{1,2} = {-1 \pm \sqrt{21}\over 2 }$$ Since $t\ge 0$ only ${-1 + \sqrt{21}\over 2 }$ is valid. Also solution to $f(t)=1-t$ are also solutions to $(*)$ so $$t^2-t-4=0$$ and so $$t_{3,4}= {1 \pm \sqrt{17}\over 2 }$$

Clearly, only ${1+ \sqrt{17}\over 2 }$ is valid. Now you can calculate both $x$...


Another way.

We have $$\sqrt{5-x}=5-x+x-x^2$$ Or $$5-x-\sqrt{5-x}-x(x-1)=0$$ or $$(\sqrt{5-x}-x)(\sqrt{5-x}+x-1)=0$$ and the rest is smooth.

Tags:

Polynomials

Functional Equations

Radicals

Algebra Precalculus

Solution Verification

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