Question

Solve for x.
x + 2(x +1 ) ^1/2 =7

Answer 1

Have you made any attempts? Or any idea on how to start?

Answer 2

I've tried squaring the whole equation to get rid of the square root but it doesn't work.

Answer 3

Hmm, did you move everything away from the \((x+1)^{1/2}\)?

Answer 4

What do you mean? Can you show it?

Answer 5

\(\Large x+2(x+1)^{1/2}=7\) into ===> \(\Large (x+1)^{1/2}=\frac{7-x}{2}\)

By subtracting

x and dividing 2

Answer 6

Yes, I did try that. What is the next step though?

Answer 7

Now you can square everything..
\(\Large ((x+1)^{1/2})^2=\frac{(7-x)^2}{(2)^2}\)

Answer 8

Leaving \(\Large x+1=\frac{1}{4}((7-x)^2)\)

Answer 9

Multiply everything by 4
\[\Large 4x+4=(49-14x+x^2)\]

\[\Large 0=45-18x+x^2\]

Answer 10

Now just factor~

Answer 11

I get it from there now. But can I also ask why didn't it work when I squared it without making (x+1)^1/2 the subject but it worked when I moved everything else to the other side?

Answer 12

to factorize just break up -18x an such a way that multiplication of coefficients gives 45 and addition gives -18...thaen you can factorize

Answer 13

Well I don't know any way of squaring \(\Large x+2(x+1)^{1/2}\).. I'm not sure if that's possible :P

Answer 14

You could change the (x+1)^1/2 into a power series but I'm not sure if you know how to do that xD

Answer 15

Oh okay, then what about solving an equation like this one: \[\sqrt{5 + x} + \sqrt{x} = 5\]
Can we still apply the same method?

Answer 16

write it as..
sqrt5+sqrtx+sqrtx=5
=>2sqrtx=5-sqrt5
now take out squre on both sides...then
4x=25+5-5sqrt5
=>4x=30-5sqrt5
=>x=30/4-5sqrt5/4=15/2-5sqrt5/4

Answer 17

@fishy13

Answer 18

But \[\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}\]