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Find all real roots of the equation

$$(x+1) \sqrt{x+2} + (x+6)\sqrt{x+7} = x^2+7x+12$$

I tried squaring the equation, but the degree of the equation became too high and unmanageable. I also tried substitutions, but it didn't work out correctly. This question was in my weekly class worksheet as were this and this question which I previously asked.

Any help will be appreciated.
Thanks.

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  • 3
    $\begingroup$ As a wild guess, given the meta hint that this was asked with the intention of an answer to be found, we try values for $x$ where $\sqrt{x+2}$ and $\sqrt{x+7}$ are "nice", preferably an integer. What square numbers are five apart from one another? $4$ and $9$. What value of $x$ would make that happen? Does that value for $x$ satisfy the equality? Could there be other integer solutions? Finding it analytically does seem to prove to be a challenge without increasing the degree dramatically. $\endgroup$ – JMoravitz May 4 '16 at 2:49
  • $\begingroup$ @JMoravitz I'm absolutely sure that an analytical solution exists (as is the case with my previous two questions). My teacher has said so. $\endgroup$ – Henry May 4 '16 at 2:53
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    $\begingroup$ Just looking for "nice" forms of terms under square root, $x=2$ is a solution. But that's just trial and error. $\endgroup$ – Kushal Bhuyan May 4 '16 at 10:12
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We have $$x^2+7x+12-(x+1) \sqrt{x+2} - (x+6)\sqrt{x+7}=0$$ Multiplying the both sides by $3$ gives $$3x^2+21x+36-3(x+1) \sqrt{x+2} - 3(x+6)\sqrt{x+7}=0\tag1$$ Now since $$\begin{align}&3x^2+21x+36\\&=x^2+5x+4+x^2+13x+42+x^2+3x-10\\&=(x+1)(x+4)+(x+6)(x+7)+(x-2)(x+5)\end{align}$$ we have, from $(1)$, $$(x+1)(x+4)+(x+6)(x+7)+(x-2)(x+5)-3(x+1) \sqrt{x+2} - 3(x+6)\sqrt{x+7}=0$$ Rearranging terms $$(x+1)(x+4-3\sqrt{x+2})+(x+6)\sqrt{x+7}\ (\sqrt{x+7}-3)+(x-2)(x+5)=0,$$ i.e. $$(x+1)\cdot\frac{(x-2)(x+1)}{x+4+3\sqrt{x+2}}+\frac{(x+6)\sqrt{x+7}\ (x-2)}{\sqrt{x+7}+3}+(x-2)(x+5)=0$$ and so $$(x-2)f(x)=0$$ where $$f(x)=\frac{(x+1)^2}{x+4+3\sqrt{x+2}}+\frac{(x+6)\sqrt{x+7}}{\sqrt{x+7}+3}+(x+5)$$ We know that $f(x)$ is positive because of $x\ge -2$.

Thus, $\color{red}{x=2}$ is the only solution.

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  • $\begingroup$ Great answer (+1), but could you provide some insight into why you did what you did? Especially rewriting as sum of 3 factors and then the rearrangement. $\endgroup$ – Aritra Das May 4 '16 at 13:53
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    $\begingroup$ @AritraDas: To have $(x-2)(\text{something positive})=0$, I wanted to make $(x-2)$s using $\sqrt{x+2},\sqrt{x+7}$, and then I noticed $\sqrt{x+7}-3=(x-2)/(\sqrt{x+7}+3)$. After these, trial and error. $\endgroup$ – mathlove May 5 '16 at 4:26
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Let $\sqrt{x+2}=a$ and $\sqrt{x+7}= b$.

So, $x + 1 = a^2 -1$

$x+6 = b^2 -1$

$ x^2 + 7x + 12= (x+3)(x+4)= (b^2-4)(a^2+2)$

Substituting and solving the system of equations.

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