Solve for $x$ : $\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9$?

Question

I want to solve the following equation for $x$ : $$\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9$$

My approach:

Let the given eq.: $$\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9 \tag {i}$$ On rearranging, we get: $$\sqrt{x-6} \, + \, \sqrt{x+6} = 9 \, - \, \sqrt{x-1} $$ On Squaring both sides, we get: $$(x-6) \, + \, (x+6) + 2 \,\,. \sqrt{(x^2-36)} = 81 + (x-1)\, - 18.\, \sqrt{x-1}$$ $$\implies 2x + 2 \,\,. \sqrt{(x^2-36)}= 80 + x - 18.\, \sqrt{x-1}$$ $$\implies x + 2 \,\,. \sqrt{(x^2-36)}= 80 - 18.\, \sqrt{x-1} \tag{ii}$$ Again we are getting equation in radical form.

But, in Wolfram app, I am getting its answer as $x=10$, see it in WolframAlpha.

So, how to solve this equation? Please help...

Answer 1

You're fine so far. Now from $(ii)$, you have

$$18 \sqrt{x-1} + 2 \sqrt {x^2-36} = 78-x.$$

You'll end up needing to square both sides a couple of more times but you'll be able to clear the radicals. Then you need to confirm that none of your potential solutions are spurious.

Answer 2

If you first think to yourself "There is no way I would be given this problem without there being a nice solution. What if all the square roots turn out to be integers?", you will find the solution fairly quickly by simple trial and error.

As for how one would try to solve, say, $$ \sqrt{x-6} + \sqrt{x - 2} + \sqrt{x+6} = 9 $$ I'd say your method of squaring and tidying up is the way to go. You go from three roots to two, and in the next step you only have one, and finally you can remove it entirely. It will be a fourth-degree equation, but with some luck there may be tricks one can do to solve it without having to use the general formula.

Answer 3

Hint:

Clearly,the LHS is increasing function of $x$

so,we cannot have multiple roots

For real solution, $x\ge6$

Also, $3\sqrt{x+6}>9>3\sqrt{x-6}$

$\implies x+6>9>x-6\iff3< x<15\implies6\le x<15$

Answer 4

Let us solve $$ \sqrt{x-6} + \sqrt{x-1} + \sqrt{x+6} = 9. $$

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Domain of solutions. It must be $x\ge 6$, $x\ge 1$ and $x\ge -6$, so $x\ge 6$. Also, $\sqrt{x-6}<\sqrt{x-1}<\sqrt{x+6}$ gives $3\sqrt{x-6}<9<3\sqrt{x+6}$ that implies: $$ {\sqrt{x-6}<3<\sqrt{x+6}}\ \Rightarrow\ {x-6<9<x+6}\ \Rightarrow\ {3<x<15}. $$ So it must be $6\le x<15$.

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We have: $$ \sqrt{x-6} + \sqrt{x+6} = 9 - \sqrt{x-1}. $$

Squaring for the first time: $$ x-6 + x+6 + 2\sqrt{x^2-36} = 81 + x-1 - 18\sqrt{x-1} $$ $$ x + 2\sqrt{x^2-36} = 80 - 18\sqrt{x-1} $$ $$ 18\sqrt{x-1} + 2\sqrt{x^2-36} = 80 - x $$

Squaring for the second time: $$ 324(x-1) + 4(x^2-36) + 72\sqrt{(x-1)(x^2-36)} = 6400 - 160x + x^2 $$ $$ 324x - 324 + 4x^2 - 144 + 72\sqrt{x^3-x^2-36x+36} = 6400 - 160x + x^2 $$ $$ 72\sqrt{x^3-x^2-36x+36} = 6868 - 484x - 3x^2 $$

Squaring for the third and last time: $$ 5184 (x^3 - x^2 - 36x + 36) = 47169424 + 234256x^2 + 9x^4 - 6648224x - 41208x^2 + 2904x^3 $$ $$ 9 x^4 - 2280 x^3 + 198232 x^2 - 6461600 x + 46982800 = 0. $$

With a bit of effort, we find: $$ (x - 10) (9 x^3 - 2190 x^2 + 176332 x - 4698280) = 0 $$ where the first factor has one real root $x=10$ and the second factor has three distinct non-integer real roots, all of them greater than $69$. We conclude that the only acceptable answer is $$ x=10. $$

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Easier method to search for an integer root, although not proving itself the unicity.

We have that $\sqrt{a}+\sqrt{b}$ or $\sqrt{a}+\sqrt{b}+\sqrt{c}$ are integers only if $a$, $b$ and $c$ are perfect squares.

Further, the sequence of differences of consecutive perfect squares $a_n=(n+1)^2-n^2$, $n\in\mathbb{N}$, is the sequence $b_n=2n+1$, $n\in\mathbb{N}$, of odd integers.

Three consecutive perfect squares, $\sqrt{x-6}$, $\sqrt{x-1}$, $\sqrt{x+6}$, whose difference are $$ (x+6)-(x-1)=6+1=7 $$ and $$ (x-1)-(x-6)=-1+6=5 $$ must be $$ {x-6=4},\quad {x-1=9},\quad {x+6=16}, $$ all of them giving $x=10$.

Answer 5

I'm not entirely sure if this approach will find all possible $x$ in any such question, nevertheless here it is giving the correct answer and is relatively short.

Since the RHS $=9$ is an integer, all three terms in the RHS under the radicals must be perfect squares. $$\therefore, x-6=p^2$$ $$x-1=q^2$$ $$x+6=r^2$$ where $p,q,r$ are integers. $$\Rightarrow x=p^2+6=q^2+1=r^2-6$$ Here we need value of only one of $p,q,r$ to find $x$

$\Rightarrow (p+q)(p-q)=-5$ $\Rightarrow (p+q),(p-q)=(-1)(5), (1,-5), (5,-1), (-5,1)$ $\Rightarrow p=±2$

Since $x=p^2+6$,

$$\therefore \fbox{x=10}$$