Quadratic equation sum from Russian book

Question

Solve $\sqrt{5-x}=5-x^2$ for $x$.

This is what I have done so far.

Method 1: \begin{align} \sqrt{5 - x} & = 5 - x^2 \\ 5 - x & = (5 - x^2)^2 \\ 5 - x & = 25 - 10x^2 + x^4 \\ 0 & = x^4 - 10x^2 + x + 20 \end{align}

Method 2: \begin{align} \sqrt{5 - x} & = (\sqrt5)^2 - (x)^2 \\ \sqrt{5 - x} & = (\sqrt5 - x)(\sqrt5 + x) \\ 1 & = \sqrt5 + x\\ x & = 1 - \sqrt5 \end{align}

Answer 1

This is good. $x^4 - 10x^2 + x + 20 = 0$

$\sqrt {5-x} = (\sqrt 5 - x)(\sqrt 5 + x)$ you cannot divide both sides as $\sqrt {5-x} \ne (\sqrt 5 - x)$

Taking a bit of a shot in the dark: $(x^2 + ax - 4)(x^2 + bx - 5) = x^4 + (-9+ab)x^2 + (-5a-4b)x - 20$

$a = -x, b = x$

$(x^2 -x - 4)(x^2 + x - 5)=0\\ x = \frac {1 + \sqrt {17}}{2},\frac {1 - \sqrt {17}}{2},\frac {-1 + \sqrt {21}}{2},\frac {-1 - \sqrt {21}}{2}$

But, we need since $\sqrt {5-x} > 0$ we require $5-x^2 > 0$ and $|x| < \sqrt 5$ eliminating 2 solutions

$x = \frac {1 - \sqrt {17}}{2},\frac {-1 + \sqrt {21}}{2}$

Answer 2

You want to solve the equation:

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

In your first attempt you squared the equation and got:

$5-x=(5-x^2)^2$

Note, that you have to be precice when you do so. The LHS is always positiv and not defined (in the real numbers) if $x>5$. The RHS can be negativ. When you just square, you create "fake solutions".

In your 2nd attempt it seems like you get

$\frac{\sqrt{5-x}}{\sqrt{5}-x}=1$ this is wrong in general.

Stick with your first try:

$x^4-10x^2+x+20=0$

One can see $x^4-10x^2+x+20=(x^2-x-4)(x^2+x-5)=0$, which reduces the problem to quadratic equations. This can be solved easily. But seeing this factorisation is not trivial and unclear. One needs a good eye or a CAS like wolframalpha.

One way would be, to suppose, that there is such a factorisation, and then go like this:

$x^4-10x^2+x+20=(x^2+ax+b)(x^2+cx+d)$

And then compare the coefficients, to get $a, b, c$ and $d$. You might get this idea, when trying to use long division and search for a linear factor, and go over the divisors of $20$.

Namely $\pm 1, \pm 2, \pm 4,\pm 5, \pm 10,\pm 20$. They all fail to give you a root.

Edit:

Taken from the discussion in the comments, I want to show a little bit more details to the comparasion of the coefficients.

After some thought process scatched before, we might write:

$x^4-10x^2+x+20=(x^2+ax+b)(x^2+cx+d)$

$(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$

Now we have:

$x^4+\color{red}{(a+c)}x^3+\color{blue}{(ac+b+d)}x^2+\color{orange}{(ad+bc)}x+\color{green}{bd}=x^4+\color{red}{0}x^3-\color{blue}{10}x^2+\color{orange}{1}x+\color{green}{20}$

This leads to the equation system:

$a+c=0$

$ac+b+d=-10$

$ad+bc=1$

$bd=20$

Answer 3

The domain gives $x\leq5$ and $-\sqrt5\leq x\leq\sqrt5$ or $$-\sqrt5\leq x\leq\sqrt5.$$ Now, let $5=y$.

Thus, $$\sqrt{y-x}=y-x^2$$ or $$y-x=y^2-2x^2y+x^4$$ or $$y^2-(2x^2+1)y+x^4+x=0$$ or $$y^2-(2x^2+1)y+\frac{1}{4}(2x^2+1)^2-x^4-x^2-\frac{1}{4}+x^4+x=0$$ or $$\left(y-\frac{2x^2+1}{2}\right)^2-\left(x-\frac{1}{2}\right)^2=0$$ or $$(y-x^2-x)(y-x^2+x-1)=0$$ or $$(5-x^2-x)(4-x^2+x)=0,$$ which with our domain gives the answer: $$\left\{\frac{\sqrt{21}-1}{2},\frac{1-\sqrt{17}}{2}\right\}.$$