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I need to find $x$, given that $$\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$$ I simplified this to $x^4+14x^3+105x^2+68x-188=0$. According to Symbolab, that is not correct. That's why I'm goint to write out my attempt so you can point out where my mistake is.

My attempt

$$\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$$ $$\left(\sqrt{x+4+2\sqrt{x+3}}\right)^2=\left(\frac{x+8}{3}\right)^2$$ $$x+4+2\sqrt{x+3}=\frac{x^2+16x+64}{9}$$ $$9x+36+18\sqrt{x+3}=x^2+16x+64$$ $$-x^2-7x-28=-18\sqrt{x+3}$$ $$x^2+7x+28=18\sqrt{x+3}$$ $$(x^2+7x+28)^2=(18\sqrt{x+3})^2$$ $$x^4+7x^3+28x^2+7x^3+49x^2+196x+28x^2+196x+784=324x+972$$ $$x^4+14x^3+105x^2+68x-188=0$$ Where is my mistake? Even if this were true, I still wouldn't be able to solve it without a calculator (I can't use Rational Root Theorem on such a big numbers!).

By the way, the solution should be (again, according to Symbolab) $x \in \{1,-2\}$.
Michael Rozenberg
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Hanlon
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    You can use the rational root theorem, which suggests that you check (among others, so why not first?) $\pm1$ and $\pm2$. Checking that $1+14+105+68-188=0$ and $16-112+420-136-188=0$ is straigtforward, so you find $1$ and $-2$ as roots. After that, you are left with a quadratic – Hagen von Eitzen Jun 19 '18 at 10:04
  • To find all possible solutions, take a look at the quartic equation problem in here: https://en.wikipedia.org/wiki/Quartic_function . – Alex Silva Jun 19 '18 at 10:09

4 Answers4

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It's $$\sqrt{1+2\sqrt{x+3}+x+3}=\frac{x+8}{3}$$ or $$\sqrt{\left(1+\sqrt{x+3}\right)^2}=\frac{x+8}{3}$$ $$1+\sqrt{x+3}=\frac{x+8}{3}$$ or $$\sqrt{x+3}=\frac{x+5}{3}$$ and since $x\geq-3$, it's $$9(x+3)=(x+5)^2,$$ which gives $x=1$ or $x=-2.$.

Michael Rozenberg
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What you did is fine. Now, you can use the rational root theorem in order to find the roots $1$ and $-2$. Since your polynomial is $(x-1)(x+2)(x^2+13x+94)$, there are no more real roots. Note however that you still must check whether or not $-2$ and $1$ are solutions of the original equation.

José Carlos Santos
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You can brutally use Ferrari formula but no need here.

Search for obvious Roots , here 1 is clearly an obvious Roots then factorise by :

$$X-1$$

And try again with obvious roots. $0;1;2;-1,i,-i$

Pagode
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  • But where is my mistake? Or is Symbolab wrong? – Hanlon Jun 19 '18 at 10:05
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    No mistake ! Just plug 1 in your équation and verify ! As I suggest in general try obvious roots to factorize – Pagode Jun 19 '18 at 10:06
  • What kind of formula is "Ferrari formula"? I can't find anything about it. – Hanlon Jun 19 '18 at 10:10
  • Think it is better known in France just look at formula there https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Ferrari – Pagode Jun 19 '18 at 10:34
  • I translate the original idea for cultural knowledge: you replace $z^4$ by $ (z^2+y^2)^2-2yz^2-y^2 $ where $y$ is a parameter to fix further then you factorize your original equation (like the expression above) and identify y. – Pagode Jun 19 '18 at 10:36
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Your question was how to determine the value of $x$ in the given equation: $$ \sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3} $$

1.) $ x+4+2 \sqrt{x+3} = \bigg( \frac{x+8}{3} \bigg)^2 $

2.) $ x+4+2 \sqrt{x+3} = \frac{x^2+8^2}{3^2} $

3.) $ 9(x+4+2 \sqrt{x+3}) = (x+8)^2 $

4.) $ 9x+36+18 \sqrt{x+3} = (x+8)(x+8) $

5.) $ 18 \sqrt{x+3}) = (x^2+16x+64)-9x-36 $

6.) $ \sqrt{x+3}) = \frac{x^2+7x+28}{18} $

7.) $ x+3 = \bigg( \frac{(x^2+7x+28)({x^2+7x+28})}{18 \times 18} \bigg) $

8.) $ x+3 = \frac{x^4+14x^3+105x^2+392x+784}{324} $

9.) $ 324x+972 = x^4+14x^3+105x^2+392x+784 $

10.) $ x^4+14x^3+105x^2+68x=188 $

The outcome of the equation is that it is a $4^ \text{th}$ degree polynomial of the form $ a_1x^4+a_2x^3+a_3x^2+a_4x+a_4=0 $

$$ x^4+14x^3+105x^2+68x-188=0 $$

The solution given by my calcuator is:

$ x_1=1, x_2=-2 $

To find out how that comes about, refer to Quartic polynomial solutions

Rhodie
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  • You reproduced the same OP calculations. There is nothing new in your answer. – Alex Silva Jun 19 '18 at 16:01
  • Then that proves the OP was correct but I went further to prove it using my own work that the answer was correct and added a reference to improve the OP's understanding of why which is exactly what was asked. – Rhodie Jun 20 '18 at 16:29