help with radicals: solve sqrt x - 2 = 5

[spicytakoyum]

SOLVE THE EQUATION: ?x-2=5

(?x-2^2)=(5)^2
(?x-2)(?x-2)=25

I understand all of it until this:

x-4?x+4=25
x-4?x=21

Where did the x-4 radical x + 4 come from?
Thank you,
Mabel

 

[Mrspi]

Re: help with radicals

SOLVE THE EQUATION: ?x-2=5

(?x-2^2)=(5)^2
(?x-2)(?x-2)=25

I understand all of it until this:

x-4?x+4=25
x-4?x=21

Where did the x-4 radical x + 4 come from?
Thank you,
Mabel

[sqrt(x - 2)]^2 = 5^2

Squaring and taking the square root are INVERSE operations; that is, one "undoes" the other. So, squaring the square root of (x - 2) should give you just (x - 2):

x - 2 = 25

Now...continue.

 

[Loren]

Re: help with radicals

Your notation is confusing. ?x-2=5 leaves us not knowing whether you mean \(\displaystyle \sqrt{x} - 2 = 5\) or \(\displaystyle \sqrt{x-2} = 5\). If you mean the former it could be left as is or written as ?(x)-2=5.
If you mean the latter, you should probably write it as ?(x-2)=5. In either case, you should probably get the radical expression all by itself on one side of the equal sign, then square both sides and solve for x. Be sure to check your result by plugging your result(s) back into the original equation to make sure it is telling the truth.

 

[Deleted member 4993]

The way you (mabel) tried show the solution of the probelm - I think it is

\(\displaystyle \sqrt{x} - 2 = 5\)

then easiest way to do the problem,

\(\displaystyle \sqrt{x} = 5 + 2\)

\(\displaystyle [\sqrt{x}]^2 = 7^2\)

\(\displaystyle x^2 = 49\)

However, your question was related to expansion of:

\(\displaystyle (a+b)^2 = a^2 + 2ab + b^2\)

You should get that FOIL-ing \(\displaystyle (a+b)^2\)

\(\displaystyle (a+b)^2 = (a+b)\cdot\ (a+b) = (a\cdot\ a) + (a\cdot\ b) + (b\cdot\ a) + (b\cdot\ b) = a^2 + 2ab + b^2\)