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

Quote:

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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

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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.