Question

6r^4+11r^2=0

Answer 1

does the equations look like this on your paper?

(6r)^4 + (11r)^2 = 0

Answer 2

No parenthesis in the equation...

Answer 3

That means you would only put the r to the 4th power and square the second r.

is 15r^6 an answer?

Answer 4

The answer is -10/2i, 10/2i -6/3, 6/3. Radical sign before the ten and six....it's a quadratic equation...tryin to find out a easy process

Answer 5

Square root sign I meant

Answer 6

\[6r^4+11r^2=0\\
r^2(6r^2+11)=0\]
\[r^2=0\text{ and }6r^2+11=0\]
The first equation tells you that r = 0 is a solution.
Solving for r in the second equation gives you
\[r=\pm \;i\sqrt{\frac{11}{6}}\]
Are you sure that's the right equation?

Answer 7

That's how it's writen in the book...

Answer 8

http://www.wolframalpha.com/input/?i=6r%5E4%2B11r%5E2%3D0