Expert: Jeff White - 10/16/2004

Question
I have 2 problems that I need help on fast. You have to use the quadratic formula to solve the following. Leave irrational roots in simplest radical form.

1. 3x squared - 4x - 5=0
2. 2x squared - 3x + 1=0

Answer
I had this all ready to send yesterday, then I hit something and deleted it, and didn't have time yesterday to re-do it.  Sorry for the delay.

You can indicate "squared" in two ways.  Most common online is "^2".  Another option is to hold down your Alt key and enter the numbers 0178...that should give you a true exponent, like this: "²".

3x² - 4x - 5 = 0
x = (4 ± sqrt((-4)² - 4(3)(-5))/2(3)
= (4 ± sqrt(16 + 60))/6
= (4 ± sqrt(76))/6
= (4 ± 2*sqrt(19))/6
= (2 ± sqrt(19))/3

2x² - 3x + 1 = 0
x = (3 ± sqrt((-3)² - 4(2)(1))/2(2)
= (3 ± sqrt(9 - 8))/4
= (3 ± 1)/4
= {1, 1/2}

Hope this helps.

Jeff