Asked by logan

i really need help on these..i don't understand ....i need to find the real number solutions of the equations...

-2v^6=16

3s^4+15s^2-72=0

3c^7=81c^4

3q^4+3q^3=6q^2+6q

any help would be appreciate...thank you!!
Answered by drwls
Rewrite as 3c^4(c^3 -27) = 0
This is true if either c=0 or c^3 = 27
27 is the cube of 3, so c=3 is the other solution.
Answered by logan
can you help me with the other three!!
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