Solve $$\sqrt{3+\sqrt{3+x}}=x$$
My try:
$$\sqrt{3+\sqrt{3+x}}=x \\ 3+\sqrt{3+x}=x^2\\\sqrt{3+x}=x^2-3\\3+x=(x^2-3)^2$$
$$x^4-6x^2+9=x+3\\x^4-6x^2-x+6=0$$
Now ?
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3You don't need absolute value at the end. $(\sqrt x)^2 = x$ and $x+3\geq 0$ anyway. – Ennar May 14 '17 at 18:45
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@Ennar. ok .... – Almot1960 May 14 '17 at 18:46
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Note that x=1 is solution of the last line but not the first. You have included extra roots in the few lines of math. – Juanito May 14 '17 at 18:48
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You sound unconvinced about Ennar's comment., while normally $(\sqrt{x})^2=|x|$, here we notice that $x=\sqrt{\text{something}}$ and $\sqrt{\text{something}}$ is always greater than or equal to zero in the case we are limiting our attention to the real numbers, so we know $x\geq 0$ and therefore $|x|=x$ – JMoravitz May 14 '17 at 18:48
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Note that with $y:=\sqrt{3+x}$, what we need is $\sqrt{3+y}=x$. – Berci May 14 '17 at 18:49
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@Arthur: No, the $x$ disappeared – Berci May 14 '17 at 18:49
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Solve the Quadratic equation – Tom Carter May 14 '17 at 18:50
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@JMoravitz, $(\sqrt x)^2=|x|$ is true, but there is no need for absolute value. (Real) square root is right inverse of $\cdot^2\colon \mathbb R\to\mathbb R_{\geq 0}$ by definition. – Ennar May 14 '17 at 18:53
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Just noticed the $x$ in the equation my bad – Tom Carter May 14 '17 at 18:56
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Similar post: https://math.stackexchange.com/q/1926750/321264. – StubbornAtom May 14 '17 at 18:58
4 Answers
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There is no need to deal with any quartic equation in $x$. The key is the function on LHS $x \mapsto \sqrt{ 3 + \sqrt{3+x}}$ is a composition of the map $x \mapsto \sqrt{3+x}$ with itself.
The map $x \mapsto \sqrt{3+x}$ is strictly increasing whenever it is defined (i.e $x \ge -3$)
If $\sqrt{3+x} > x$, then $\sqrt{3 + \sqrt{3+x}} > \sqrt{3+x} > x$.
If $\sqrt{3+x} < x$, then $\sqrt{3 + \sqrt{3+x}} < \sqrt{3+x} < x$.
If we want $\sqrt{3 + \sqrt{3+x}} = x$, we need $\sqrt{3+x} = x$. This leads to $$x^2 = 3 + x \iff x^2- x - 3 = 0 \implies x = \frac{1 \pm \sqrt{13}}{2}$$
Since $x = \sqrt{3 + \sqrt{3 + x}}$ is supposed to be non-negative, $x = \frac{1 + \sqrt{13}}{2}$ is the only possible solution (and it does work).
achille hui
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$$\sqrt{3+x}=x^2-3$$ $$3+x=(x^2-3)^2$$ $$x^4 -6x^2 +9 -x - 3=x^4 -6x^2-x+6=0$$ $$(x-1)(x+2)(x^2-x-3)=0$$
None of $1, -2, \frac{1-\sqrt{13}}{2}$ are not answers, due to our double-squaring. Basically, we need the following inequalities to be true:
$$x+3 \ge 0$$ $$x\ge 0$$ $$\sqrt{3+x} = x^2 - 3 \ge 0$$
Thus $x \ge \sqrt{3}$, and $x=\frac{1+\sqrt{13}}{2}$
Jay Zha
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1And now, the false solutions created by the successive squarings are ... and the unique true solution is... – Did May 14 '17 at 18:49
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@Almot1960, you guess the integer solutions and then do polynomial division. It's easy to guess them since they must divide $6$. Please see this. – Ennar May 14 '17 at 18:55
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@Did good point, it's worth to dig in to the final answer, instead of stopping there. – Jay Zha May 14 '17 at 19:01
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@Almot1960 yea, I mostly observed and tried. Did not follow any methodology to do the factorization. – Jay Zha May 14 '17 at 19:03
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First state the conditions that $x$ has to satisfy:
- The initial equation requires $x\ge 0$.
- $\sqrt{3+x}=x^2-3$ requires $x^2\ge 3$, i.e., taking into account the previous condition, $x\ge \sqrt3$. The resulting equation is easy to factorise: rewrite it as \begin{align} x^4-x-6x^2+6&=x(x^3-1)-6(x^2-1)=(x-1)\bigl(x(x^2+x+1)-6(x+1)\bigr)\\&=(x-1)(x^3+x^2-5x-6). \end{align} One tests the existence of rational roots for the second factor, among $\pm1, \pm2,\pm3,\pm6$, one finds $-2$, so dividing by $x+2$: $$x^4-x-6x^2+6=(x-1)(x+2)(x^2-x-3).$$ Set $p(x)=x^2-x-3$. Its discriminant is $\Delta=13$. As $p(\sqrt 3)=-\sqrt 3<0$, $\sqrt 3$ separates the roots, so the only root $\ge\sqrt3$ is $$x=\frac{1+\sqrt{13}}2.$$
Bernard
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We have to factorize $x^4 - 6x^2 - x + 6 = 0.$
We would observe that the values of x would be $1,-2,\frac{1±\sqrt{13}}{2}$.
Haran
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