Question

How do I use the quadratic formula to solve `125x^3 - 1 = 0`?

Answer 1

Hopefully you know what "difference of cubes" is, since this problem involves this method before you can use the quadratic formula... You'll see what I mean.

To begin, because you are subtracting you know that the difference of cubes should be in the form `(x^3 - y^3) = (x - y)(x^2 + xy +y^2)`.

To make this problem simpler, get your given equation in the form `(x^3 - y^3)` . You should know that `5^3 is 125` and `1^3` is `1` ... so rewrite the problem as follows:

`(5x)^3 - (1)^3`

Now, you know that (5x) is the x value of the difference of cubes formula, and that 1 is the y value. So, sub these values into the formula (x - y)(x^2 + xy +y^2) like so:

`= (5x - 1)[(5x)^2 + (5x)(1) + (1)^2]`

Simplify it all:

`(5x - 1) (25x^2 + 5x + 1)`

Now, because the bracket on the left is in linear form (just x to the power of 1), you can solve it:

5x - 1 = 0
5x = 1
x = 1/5

That's your first solution. Now, you use the quadratic formula to solve the other bracket, since it is quadratic.

Recall the quadratic formula is:

Now sub in all your values from 25x^2 + 5x +1. a = 25 , b = 5 , and c = 1 .

Now solve: WAIT! If you check the discriminant (under the root sign), you'll notice that it comes out to be negative ! This means there are no real roots! If the discriminant was NOT NEGATIVE, you would solve for the roots as usual and you would have 3 roots!

As such, the one root you got before (x=1/5) is the only real root! Hopefully you understood all this and hopefully I was of some help!