SOLUTION: We are solving quadratic equation. I tried to sole this problem, but I'm having some trouble. Please help me solve this quadratic equation: {{{8x^3-1=0}}}. Any help will be appreci

Question 80633This question is from textbook
: We are solving quadratic equation. I tried to sole this problem, but I'm having some trouble. Please help me solve this quadratic equation: . Any help will be appreciated. Thank you,
Jennie
This question is from textbook

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
solve this quadratic equation: 8x^3-1 = 0
:
You should recognize this as "the difference of cubes" which come under a section
referred to as "Special Factoring Formulas", given in a book that I have as:
:
u^3 - v^3 = (u - v)(u^2 + uv + v^2)
:
in your problem u^3 = 8x^3 and -v^3 = -1
:
So we have:
(8x^3 - 1) = (2x - 1)(4x^2 + 2x + 1)
:
You check this for yourself: multiply (2x-1) * (4x^2 + 2x +1) you should get the
original equation
:
Look in the index of your algebra book under "difference of cubes" you should see the special factor formulas there
:
-----------------------------------------------------------------
Added after receiving inquiry from student:
(2x-1)(4x^2 + 2x + 1) = 0
The 1st factor:
2x - 1 = 0
2x = +1
x = 1/2
:
The other factor: (4x^2 + 2x + 1),= 0, has no real roots The discriminant
of the equation is b^2 - 4*a*c, as I am sure you know.
In this case, a=4; b=2; c=1: 2^2 - 4*4*1 = -12, remember when the
discriminant is negative it has no real roots
:
A graph would look like this:

:
Note that it crosses the x axis only at +.5

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