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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}

Bladewood
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L Lawlit
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3 Answers3

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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:

$a+c=0$

$ac+b+d=-10$

$ad+bc=1$

$bd=20$

Cornman
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    I was just typing what you just added ! Cheers and $+1$. – Claude Leibovici Aug 07 '18 at 04:35
  • Sir how did u reduce it to (x²-x-4) and (x²+x-5) – L Lawlit Aug 07 '18 at 04:42
  • Coz there is no cube in the sum. And that's why we can't divide x² throughout the sum. When we divide throughout by x² I am getting the two roots u said. – L Lawlit Aug 07 '18 at 04:44
  • @KalpitPrabhu I gave a method in my last edit. You compare the coefficients. Factor out the general form $(x^2+ax+b)(x^2+cx+d)$ and solve for a, b, c and d. Then you get to this form. – Cornman Aug 07 '18 at 04:45
  • @KalpitPrabhu I do not understand your second comment. – Cornman Aug 07 '18 at 04:46
  • a is 1. b is -10. c is 1. d is 20. Right???? – L Lawlit Aug 07 '18 at 04:50
  • Is it a formula????. Coz never heard this equation before – L Lawlit Aug 07 '18 at 04:51
  • No. You have to factor it out. We have $(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$. Hence it has to be: $a+c=0$, $ac+b+d=10$, $ad+bc=1$ and $bd=20$. This is the comparision of the coefficients. – Cornman Aug 07 '18 at 04:54
  • @KalpitPrabhu No, it is not a formula. It is a general method. It is a little bit advanced, but can be solved with elementary solutions. Normally it results in solving an equation system. As shown in the comment above. – Cornman Aug 07 '18 at 04:55
  • @KalpitPrabhu I added some more details in my answer. I hope the method gets clear. – Cornman Aug 07 '18 at 05:05
  • Sir I swear but I am not being able to understand it – L Lawlit Aug 07 '18 at 11:45
  • @KalpitPrabhu Where do you struggle? – Cornman Aug 07 '18 at 11:45
  • Can u show in detail please. – L Lawlit Aug 07 '18 at 11:45
  • @KalpitPrabhu Have you seen the edit? – Cornman Aug 07 '18 at 11:46
  • I am not understanding the part after x⁴-10x²+x+20 – L Lawlit Aug 07 '18 at 11:48
  • Yes sir but not understanding it – L Lawlit Aug 07 '18 at 11:50
  • @KalpitPrabhu I tried to explain it as good as I could. The idea is, that we want to factorise the polynomial, since we have no formula for this type of equation. Thats why we want to write it as product of two quadratic polynomials. This can then be easily solved with the pq-formula. But we have to find these two quadratic polynomials first. The form of them is $x^2+ax+b$ and $x^2+cx+d$. And when we multiply them, we have to get $x^4-10x^2+x+20$. Thats why it is $x^4-10x^2+x+20=(x^2+ax+b)(x^2+cx+d)$. After factoring that out, we are left with solving the equation system, shown above. – Cornman Aug 07 '18 at 11:54
  • Sir what is the value of a, b, c, d. That's where I am getting confused. And sir how u got x³ – L Lawlit Aug 07 '18 at 12:04
  • You can write $x^4-10x^2+x+20=x^4+0\cdot x^3-10x^2+x+20$. For solving the system of equations, you can start with the first $a+c=0$. This gives $c=-a$ immediatly. Now we use that in the 3rd equation and get: $ad-ab=1\Leftrightarrow a(d-b)=1$. Hence $d=b+1$. Taking in count, that $db=20$ we can see, that $d=4$ or $d=-5$ and $d=5$ or $d=-4$. I leave the rest to you. – Cornman Aug 07 '18 at 12:20
  • What to do after we get d??. I am now being able to understand the solution – L Lawlit Aug 07 '18 at 13:08
  • @KalpitPrabhu You solve for the remaining coefficient. Which is $b$. – Cornman Aug 07 '18 at 14:08
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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: $$\left\{\frac{\sqrt{21}-1}{2},\frac{1-\sqrt{17}}{2}\right\}.$$

Michael Rozenberg
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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

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

Doug M
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