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

1 Answer

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

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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!