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Number of real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is

Plan

Put $x^2+6x+7=f(x)$. Then i have $f(f(x))=x$

For $f(x)=x$

$x^2+5x+7=0$ no real value of $x$

For $f(x)=-x$

$x^2+8x+7=0$

$x=-7,x=-1$

Solution given is all real solution

Help me please

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    $\begingroup$ There are no real roots. $\endgroup$ – Servaes May 21 at 10:33
  • $\begingroup$ Maybe useful: $$ \left( {x}^{2}+6\,x+7 \right) ^{2}+6\,{x}^{2}+35\,x+49= \left( {x}^{2 }+5\,x+7 \right) \left( {x}^{2}+7\,x+14 \right).$$ $\endgroup$ – user0410 May 21 at 10:47
  • $\begingroup$ $f(x)=-x$ does not help because $f(-x)\neq -f(x)$. $\endgroup$ – Jyrki Lahtonen May 21 at 14:39
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Let $x^2+6x+7=y$.

Thus, $y^2+6y+7=x$, which gives $$x^2+6x+7-(y^2+6y+7)=y-x$$ or $$(x-y)(x+y+7)=0.$$ Thus, $x=y,$ which does not give real solutions, or $$y=-x-7,$$ which gives $$x^2+6x+7=-x-7.$$ Can you end it now?

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You are right that $f(f(x))=x$

This gives $f(x)=f^{-1}(x)$

Do you know anything about the relationship between a function and its inverse?

Try sketching both $y=f(x)$ and $y=f^{-1}(x)$ on the same axes...

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On completing the square, we have \begin{align} x= (x^2+6x+7)^2+6(x^2+6x+7)+7 &= ((x+3)^2-2)^2+6((x+3)^2-2)+7\\ &= (x+3)^4-4(x+3)^2+4+6(x+3)^2-5\\ &= (x+3)^4+2(x+3)^2-1= ((x+3)^2+1)^2-2 \geq 1-2=-1 \end{align} Further, we alsoknow that $(x+3)^2$ is monotonically increasing for $x\geq -3$. Thus, we get that \begin{align} x= ((x+3)^2+1)^2-2\geq ((-1+3)^2+1)^2-2 = 23. \end{align} Further, $((x+3)^2+1)^2-2\geq 3^34x+3^4-2\geq x, \forall x\geq 23$.

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  • $\begingroup$ The fact that there are no positive roots is almost obvious. What about negative roots? $\endgroup$ – Kavi Rama Murthy May 21 at 11:59
  • $\begingroup$ We have already got that $x>23$. So, no need to worry about negative roots! $\endgroup$ – Geethu Joseph May 21 at 16:56
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Set $y=x^2+6x+7,$ and note that the resulting system $$y^2+6y+7-x=0,\,\,\,x^2+6x+7-y=0$$ is symmetric. That is, performing the transformation $(x,y)\mapsto(y,x)$ leaves the system unchanged except for a permutation. Thus, for every solution $(x,y),$ it is always the case that $(y,x)$ is also a solution, where $x,y\in\mathrm R.$

It follows that this system can only have an even number of real solutions, which must be in pairs that are reflections of each other in the line $y=x,$ and they cannot both fall on this line because the given equation is a quartic with real coefficients, and as such cannot have an odd number of real roots.

But the parabolas defined by the system above are orthogonal, congruent parabolas. Thus, they can only intersect along the line $x=y.$ This contradicts the observation above that the points of intersection cannot fall on this line. Thus, there are no real solutions for the system, and therefore to the original quartic.

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Upon completing the square the equation $$(x^2+6x+7)^2+6(x^2+6x+7)+7=x$$

$$(x^2+6x+7)^2+6(x^2+6x+7)+9=x+2$$

$$[(x^2+6x+7)+3]^2=x+2$$

Which simplifies to $$[(x+3)^2+1]^2=x+2$$ which has no real solutions.

Upon comparison of derivatives for $x>-2$, we see that function on the LHS is always above the one on the RHS so they do not meet at any real values.

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  • $\begingroup$ +1 Why the downvote? $\endgroup$ – Servaes May 21 at 13:39
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Rearrange: $$(x^2+6x+7)^2+6(x^2+6x+7)+7=x \iff \\ ((x+3)^2-2)^2+6((x+3)^2-2)+10=x+3 \iff \\ (x+3)^4+2(x+3)^2-(x+3)+2=0 \iff \\ (x+3)^4+(x+3-\frac12)^2+(x+3)^2+\frac74=0.$$ The LHS is positive, hence, there is no real solution.

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