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Calculate the real solutions $x\in\mathbb{R}$ to

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

My Attempt:

We know that $5-x\geq 0$ and thus $x\leq 5$ and

$$ \begin{align} 5-x^2&\geq 0\\ x^2-\left(\sqrt{5}\right)^2&\leq 0 \end{align} $$

which implies that $-\sqrt{5}\leq x \leq \sqrt{5}$. Now let $y=\sqrt{5-x}$. Then

$$ \tag1 y^2=5-x $$

and the equation converts into

$$ \begin{align} y &= 5-x^2\\ x^2 &= 5-y\\ y^2-x^2 &= 5-x-(5-y)\\ y^2-x^2 &= y-x\\ (y-x)(y+x)-(y-x) &= 0\\ (y-x)(y+x-1) &= 0 \end{align} $$

So either $y=x$ or $x+y=1$.

Case 1 ($y=x$):

We can plug this into $(1)$ to get

$$ \begin{align} y^2 &= 5-x\\ x^2 &= 5-x\\ x^2+x-5 &= 0\\ x &= \frac{-1\pm \sqrt{1+20}}{2} \end{align} $$

Since $-\sqrt{5}\leq x\leq \sqrt{5}$, the only solution is

$$ x = \frac{-1+\sqrt{21}}{2} $$

Case 2 ($y=1-x$):

We can plug this into $(1)$ to get

$$ \begin{align} y^2 &= 5-x\\ (1-x)^2 &= 5-x\\ 1+x^2-2x &= 5-x\\ x^2-x-4 &= 0\\ x &= \frac{1\pm\sqrt{17}}{2} \end{align} $$

Since $-\sqrt{5}\leq x\leq \sqrt{5}$, the only solution is

$$ x = \frac{1-\sqrt{17}}{2} $$

So final solution is

$$ x \in \left\{\displaystyle \frac{1-\sqrt{17}}{2}, \frac{-1+\sqrt{21}}{2} \right\} $$

Is it possible to solve this problem using geometry? For example, could we use the properties of a half-circle and a parabola?

Fabrosi
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juantheron
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  • How will geometry get you your square roots? Certainly there will be some clever geometric arrangement that would appear to give the answer, but the amount of work to prove it correct would be at least as much as the direct algebraic solution. So what is the point? – user21820 Aug 11 '15 at 11:43
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    Graphing both sides of the equation gives you a sideways half-parabola (the LHS) and an upwards parabola (the RHS), with two intersection points. Like others here, I don't see the point of doing the problem that way. – Rory Daulton Aug 11 '15 at 11:58

2 Answers2

5

$$ \begin{align} \sqrt{5-x}&=5-x^2\\ 5-x &= \left(5-x^2\right)^2\\ 5-x &= x^4-10x^2+25\\ x^4-10x^2+25-5+x &= 0\\ x^4-10x^2+x+20 &= 0\\ (x^2-x-4)(x^2+x-5) &= 0 \end{align} $$

$$ \begin{align} x^2-x-4=0 &\vee x^2+x-5=0\\ x=\frac{-(-1)\pm\sqrt{(-1)^2-4\cdot 1 \cdot (-4)}}{2\cdot 1} &\vee x=\frac{-1\pm\sqrt{1^2-4\cdot 1 \cdot (-5)}}{2\cdot 1}\\ x=\frac{1\pm\sqrt{17}}{2} &\vee x=\frac{-1\pm\sqrt{21}}{2} \end{align} $$

Two of the $4$ solutions are good:

$$ \begin{align} x_1 &= \frac{1-\sqrt{17}}{2}\\ x_2 &= \frac{-1+\sqrt{21}}{2} \end{align} $$

I don't see the positive thing in using geometry! This is the fastest way.

Fabrosi
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Jan Eerland
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    That is just another (IMO easier) way to get the solutions via algebraic means, OP asked for geometric interpretation/solution. – Hirshy Aug 11 '15 at 11:52
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    @Hirshy: Sometimes, the best answer is to tell the asker that the question is not ideal and that there is a better approach. – Rory Daulton Aug 11 '15 at 11:55
  • @RoryDaulton I commented when there was just the plain algebraic way with no comment on geometry whatsoever. And I agree that this solution is better and faster than the one provided in the question (although getting to $$(x^2-x-4)(x^2+x-5)$$ might be tricky when you're not used to factorizing polynomials like this). – Hirshy Aug 11 '15 at 12:00
3

You can see the solutions as the abscissa of the intersection points of the two curves $y=5-x^2$ (a parabola) and $y=\sqrt{5-x}$ (upper half of a parabola). These twoo parabolas intersect at 4 points, but only two of them lie in the $y>0$ half-plane.

Intelligenti pauca
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