I cant make this problem work for whatever the reason is.

(2z)^(1/2)-(z+7)^(1/2)=1

For whatever reason i want to square all of it but then one i solve it down i get -6 and i know thats not right.

thanks to anyone

AC

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- Aug 28th 2008, 06:10 PMCasas4Algebra Review
I cant make this problem work for whatever the reason is.

(2z)^(1/2)-(z+7)^(1/2)=1

For whatever reason i want to square all of it but then one i solve it down i get -6 and i know thats not right.

thanks to anyone

AC - Aug 28th 2008, 06:18 PMJOhkonut
- Aug 28th 2008, 06:36 PMWeezygraphing problem
what is wrong with the graph to this equation?

+- the square root of 1-x^2 - Aug 28th 2008, 08:05 PMCasas4what do i do from here
Ok so when i square root both sides do i square root (1+(z+7)^1/2 or just (z+7)^1/2

When i tried doing it (1 +((z+7)^1/2))^2 i still couldnt get it to work.

Sorry for whatever reason i cant make this problem work. - Aug 28th 2008, 08:54 PMJOhkonut
You're never going to square root anything in this problem. You're squaring both sides. For (1 +((z+7)^1/2))^2,

which is the same thing as $\displaystyle (1+\sqrt {z+7})^2$, you have to use FOIL by splitting the square into 2 binomials. $\displaystyle (1+\sqrt {z+7})(1+\sqrt {z+7})$ FOIL is the order of what terms to multiply when multiplying 2 binomials.

First - Multiply the first terms ($\displaystyle 1$ and $\displaystyle 1$)

Outer - Multiply the outer terms ($\displaystyle 1$ and $\displaystyle \sqrt {z+7}$)

Inner - Multiply the inner terms ($\displaystyle \sqrt {z+7}$ and $\displaystyle 1$)

Last - Multiply the last terms ($\displaystyle \sqrt {z+7}$ and $\displaystyle \sqrt {z+7}$) - Aug 28th 2008, 10:07 PMmr fantastic
Do not post new questions in existing threads. Start a new thread.

Do not ask for help in Live Chat. Live Chat (as it clearly states) is for small talk. (Against my better judgement I have edited an answer into your chat).

Your question here makes no sense - it requires more information.