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

Question

So I tried to solve this problem for x

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?

When you square the difference of the two radicals, your middle term is incorrect. — , Jul 20 '14 at 01:35
@WebMatser, Related : http://math.stackexchange.com/questions/400403/solve-sqrtx4-sqrtx1-1-for-x — lab bhattacharjee, Jul 20 '14 at 05:10
ah I really tried looking for related posts first, but I didn't see that one — Gᴇᴏᴍᴇᴛᴇʀ, Jul 21 '14 at 01:38

score 9 · Answer 1 · answered Jul 20 '14 at 01:43

9

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.

answered Jul 20 '14 at 01:43

André Nicolas

507,029

1

Wicked method! I wouldn't have thought to invert and rationalise. – Khallil Jul 20 '14 at 01:57
Nice result! If we "subtract" the original equation and divide the result by 2, we get $\sqrt{x+7}=2$. – mike Jul 20 '14 at 02:37
Yes, I wondered about writing it up that way, the numbers are smaller. But subtraction is "more complicated" than addition. – André Nicolas Jul 20 '14 at 02:43

score 4 · Answer 2 · answered Jul 20 '14 at 01:48

4

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

answered Jul 20 '14 at 01:48

Mathsource

5,393

This method is better than direct squaring of the original equation. – mike Jul 20 '14 at 02:36

score 3 · Accepted Answer · answered Jul 20 '14 at 01:59

3

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.

answered Jul 20 '14 at 01:59

Zarrax

44,950

What was the second mistake I made? Just wondering – Gᴇᴏᴍᴇᴛᴇʀ Jul 20 '14 at 02:17
you have $-(x+7)$ on the right instead of $x + 7$ – Zarrax Jul 20 '14 at 02:20

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

3 Answers3