Question

Solve the following equations by factoring.

Answer 1

\[(t+4)^\frac{2}{3}+2(t+4)^\frac{8}{3}=0\]

Answer 2

I am not sure if i am right, cause i met with some problems along the way, but nevermind i write my answer first.

\[\left( t+4 \right)^{\frac{2}{3}}= - 2\left( t+4 \right)^{\frac{8}{3}}\]
Then you cube both sides,
\[\left( t+4 \right)^{2} = -8\left( t+4 \right)^{8}\]
Divide (t+4)^2 throughout the equation ,
\[1=-8\left( t+4 \right)^{6}\]
\[-\frac{1}{8} =\left( t+4 \right)^{6}\]
Now here is the problem, for an even power like 6, you wont get a negative number, unless this question has something to do with complex numbers. If you dont know what complex numbers is, then im afraid i cant help you. But i assume it is, so i carry on...

\[\frac{i}{\sqrt[6]{8}}= t+4\]
\[t= \frac{i}{\sqrt[6]{8}} -4\]

Answer 3

You give me a medal, so i am right?

Answer 4

I am checking it one moment.

Answer 5

What is the number before the radical of 8?

Answer 6

6

Answer 7

Ok i thought so i was just making sure and yes i think it is correct great job.

Answer 8

Np!

Answer 9

I have one more if you would like to help me solve it also.

Answer 10

can np post it

Answer 11

Ok 1 sec i will.

Answer 12

\[5y^\frac{11}{3}+3y^\frac{8}{3}-2y^\frac{5}{3}\]

Answer 13

The one beside 3y is 8/3 and by 2y it is 5/3.

Answer 14

and what is all of that equals to? if not i cant solve it.

Answer 15

oh it is =0 sorry i forgot to put it.

Answer 16

Okay this question is slight more confusing, wait a bit while i tediously type everything out.

Answer 17

Alright sounds good... i am gonna go post some other problems while you are working on this one lol.

Answer 18

First factorise the whole equation.
\[y ^{\frac{5}{3}}\left( 5y ^{6}+3y^{3}-2 \right)=0\]
Now u realise the expression in the bracket got the form of a
quadratic equation x^2 +x+c=0
take y^3 as x to solve it.

Thus you will get,
\[y ^{\frac{5}{3}}\left( 5y ^{3} -2\right)\left( y ^{3} +1\right)= 0\]
Thus,

\[y ^{\frac{5}{3}}=0 . thus, y=0\]
or
\[5y ^{3}=2. thus, y= \sqrt[3]{\frac{2}{5}}\] next to the 2/5 radical is a 3.
or
\[y ^{3}=-1. Thus, y=-1\]