# Solving for the length of a side of a triangle

I have a problem in which I'm supposed to solve for the length of the two sides of the triangle below. I assumed that it would simply boil down to $x+5=\sqrt{4x+52}$, and converted to standard form, $-3x^2+(25-52^2)=0$, but that gives me a weird answer (the answer should be a whole number). How would I solve this?

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Proof by laziness: Probably they are asking for an integer, so let's try x=1,2,3, hm $x=3$ works, $3+5 = \sqrt{4\times 3+ 52}$. For your calculations, as you wrote it, you made an error when raising both sides to the power $2$. You shouldn't have a $52^2$. – Anon Jan 21 '13 at 1:56
You have the right idea, just the wrong form. $x+5=\sqrt{4x+52}$ means $x^2+10x+25=4x+52$, or $x^2+6x-27=0$. The positive solution to that equation will be integral. – John Moeller Jan 21 '13 at 1:56
Being this lazy, you may miss the other (possible) solutions in some cases, that is another story. – Anon Jan 21 '13 at 1:58

Since this is an isosceles triangle, we have $$x+5=\sqrt{4x+52}=2\sqrt{x+13}.$$ Squaring both sides yields $$x^2+10x+25=4x+52.$$ Moving everything over to one side, we have $$x^2+6x-27=0.$$ We can factor the left-hand side to get $$(x+9)(x-3)=0,$$ so the only possible solutions are $x=-9$ or $x=3$. Since we have lengths, the solution must be $x=3$.

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The funny thing is that $x=3$ gives a "triangle" with sides 8,8, and 16. – Anon Jan 21 '13 at 2:00

I think your only error was when you squared both sides: you ended up squaring $\,x\,$ and squaring $\,52\,$though it was under the radical, so $$(\sqrt{4x + 52}\,)^2 = 4x + 52$$ (where you obtained $4x^2 + 52^2$).

Otherwise your solution process is "spot on."

Starting over, following your correct logic:

We have an isosceles triangle (because the sides opposite of congruent angles in a triangle are equal in length), so we have the equality $$x+5=\sqrt{4x+52}$$

Squaring both sides of the equation gives us $$x^2+10x+25=4x+52.\tag{* squared}$$

Moving all the terms to one side, leaving $\,0\,$ on the left, gives us $$x^2+6x-27=0.$$

Factoring the left-hand side gives us $$(x+9)(x-3)=0,$$ so the roots (solutions) are $x=-9$ and $x=3$. Since we must have positive length for a side, the solution must be $x=3$. (The extra solution is only apparent: it was introduced when above @ (* squared)).

Hence, the length of the two "unknown" sides become $$3 + 5 = \sqrt{3\cdot 4 + 52} = \sqrt{64} = 8.$$

Observation: In fact, a "triangle" with two sides of length $8$ and the third side of length $16$ is a degenerate triangle whose vertices all lie on a line of length $16$: the vertices being the endpoints and the midpoint of that line!)

Another way to solve this is to note that the altitude ($h$) from the base divides the triangle into two congruent right triangles, with base leg length $8$: Using the pythagoean theorem: $$h = \sqrt{(x+5)^2 - 8^2} = \sqrt{(4x+52) - 8^2} \implies (x+5)^2 - 64 = 4x + 52 - 64$$ $$\implies (x+5)^2 = 4x + 52 \implies \cdots \implies x = 3$$

But your method is clearly the most straightforward!

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This answer is exactly the same as Clayton's. – John Moeller Jan 21 '13 at 2:09