Thread: One more quadratic equation

Ok, I'm on my last problem and stuck again.

16r^3-9r=0

There's not an example in my textbook (chapter 6.6) to use as a guide (we're using "Introdutory Algebra, 8th edition by Lial, Hornsby, amd McGinnis).

Any assistance or pointers to help would be appreciated.

Thanks,
Thomas

Originally Posted by tbradnc

Ok, I'm on my last problem and stuck again.

16r^3-9r=0

There's not an example in my textbook (chapter 6.6) to use as a guide (we're using "Introdutory Algebra, 8th edition by Lial, Hornsby, amd McGinnis).

Any assistance or pointers to help would be appreciated.

Thanks,
Thomas

Take out a common factor of r to give

r(16r^2-9) = 0

This gives r = 0 as a factor. You can then solve the other part by using the difference of two squares:



In your case a = 4r and b = 3

Of course... thanks. As soon as you pointed it out I could see it.

Hoping this will all come together for me at some point.