$\ \sqrt{x+39}-\sqrt{x+7}=4 $

Question

So I tried to solve this problem for x

$\ \sqrt{x+39}-\sqrt{x+7}=4 $

I multiplied both sides ($\ \sqrt{m}\cdot\sqrt{n}=\sqrt{mn} $)

$\ (\sqrt{x+39}-\sqrt{x+7})^2=16 $

$\ (x+39)-2(x^2+46x+273)-(x+7) $

$\ 0x+32+(-2x^2-92x-546) $

$\ -2x^2-92x-514 $

divide the 2 out

$\ x^2+46x+257=-8 $

$\ x^2+46x+265=0 $

Use the quadratic formula (or scientific calculator) and the answers are -6.752 and -39.248. I know the answer is exactly -3. What went wrong?

Answer 1

You made two mistakes when squaring. The correct equation is $$(x + 39) - 2\sqrt{(x+39)(x+7)} + (x + 7) = 16$$ This can be rearranged as $$2x + 30 = 2\sqrt{(x+39)(x+7)}$$ Dividing by 2: $$x + 15 = \sqrt{(x+39)(x+7)}$$ Square again $$x^2 + 30x + 225 = (x+39)(x + 7)$$ $$= x^2 + 46x + 273$$ So your equation is the same as $$30x + 225 = 46x + 273$$ So $16x = -48$ and therefore $x = -3$. When squaring equations always plug in your answer to make sure it's the actual solution. And in this case it is.

Answer 2

One can square twice. I prefer to invert and obtain $$\frac{1}{\sqrt{x+39}-\sqrt{x+7}}=\frac{1}{4},$$ and then by rationalizing the denominator get $$\frac{\sqrt{x+39}+\sqrt{x+7}}{32}=\frac{1}{4},$$ or equivalently $$\sqrt{x+39}+\sqrt{x+7}=8.$$ Now "add" to the original equation, divide by $2$. We get $\sqrt{x+39}=6$, and now it's over.

Remark: In the original post, the first step used was to square. This to some degree complicates things, the middle term should have been $-2\sqrt{(x+39)(x+7)}$. Rearrangement and squaring now get us to where we want.

Answer 3

$$ \sqrt{x + 39} - \sqrt{x + 7} = 4 \quad \Rightarrow \quad (\sqrt{x + 39})^2 = (\sqrt{x + 7} + 4)^2 \quad \Rightarrow $$ $$ x + 39 = x + 7 + 8\sqrt{x + 7} + 16 \quad \Rightarrow \quad 8\sqrt{x + 7} = 16 \quad \Rightarrow \quad x = -3 $$