Can anyone figure out these two algebra problems?

Do your own homework. It’ll teach you better than getting the answers from the internets.

Maybe you and cornbird can study together.

Why don’t you post your attempts at solving these problems so we can show you where you went wrong.

@Austinlad lol

I honestly don’t know what the answer is…sorry
BUT
You could ask friends in the same class? Or check your notes/textbook

Just tell the teacher if you really can’t solve it…

Post your work and where you get confused.

By the way, everyone gets lurve from me for not doing his homework for him

Try wolframalpha.com—solve <equation> should give you some good tips.

And, amusingly, the answer is 3.

I have enough of my own homework to complete today.

Try this website.

http://www.justanswer.com/account/sign-in.aspx?tqid=36865112

I thought homework questions were for Sunday night?

42

x^2+4^2=(x+2)^2
x²+16=x²+ 4x + 4 |-x²
16=4x+4 |-4
12=4x |:4
3=x

36╥=4/3╥x³ |x ¾
27pi=pi x³ |:pi
27=x³ |3rd sqr
3=x

no guarantees

@ragingloli: yep, both are three.

@rentluva5256 have anything else?
maybe history or chemistry?

For problem number one, you solve for the exponents. So 4^2 becomes (4×4) which is 16. So the left side of the problem is x^2 + 16. For the right side, you distribute the exponent of ^2. So (x + 2) becomes x^2 + 4. Then you add the like terms (like terms are the x^2’s and the numbers without variables and exponents). You subtract a x^2 from the right side and add the negative x^2 to the left and the positive cancels out the negative. Then you subtract four from the right side and add its negative to the 16. You get 12=0. So the first problem is prime.

@rentluva5256 for the second problem, 36╥=4/3╥x^3 is that.. 36pi=four/thirds pi x (the variable)^3?

@Finley ”So (x + 2) becomes x^2 + 4.”
Sorry, but that is wrong. It would be right if the inside of the bracket was a multiplication, but it is an addition, so you can not resolve the bracket like that.
(x+2)² is (x+2) x (x+2), after dissolving brackets you end up with x² + 2x + 2x + 2², ergo x² + 4x + 4.

try math.com

Are you sure that you wrote out the first equation correctly? Second prob: divide both sides by pi, multiply both sides by 3, divide both sides by 4, cube root, QED.

THANK YOU TO ALL!!!!! It really did help, so thank you. and @NKH12 , no. I’m sorry. That’s all I’ve had trouble in. Thanks though!

nevermind.