Expert: Richard J. Raridon - 10/3/2006

Question
Hi, I'm stumped on these questions. I just don't get these problems.
Could you please Help?

For this one it wants me to solve by completing the square.

x^2 - x - 1 = 0

(For the following Q's) Use the quadratic formula to solve the
following. Leave irrational roots in the simplest radical form.

3x^2 - 4x - 5 = 0

2x^2 - 3 = 1 = 0

This once is a word question.

"In the equation ax^2 + bx + c = 0, the value of b^2 - 4ac is called
the________________ of the quadratic equation. What does this value tell
you about the real roots of the equation?"

And last but not lest,

"Without drawing the graph of the given equation, determine

(a) how many x-intercepts the parabola has
(b) whether it's vertex lies above or below on the x-axis.

y = x^2 - 5x + 6

y = -x^2 + 2x - 1


If you could help me i know it's a lot but i've gone over these so many
times it's not funny.  

Answer
x^2 -x = 1
x^2 -x +1/4 = 1+1/4
(x-1/2)^2 = 5/4
x-1/2 = +/-(1/2)5^1/2
x = 1/2 +/-(1/2)5^1/2
Surely you have learned the quadratic formula
x = [-b +/-(b^2-4ac)^1/2]/(2a)
so for 3x^2 -4x -5 = 0, you have
x = [4 +/-(16+60)^1/2]/6 = [2 +/-(19^1/2)]/3
I presume you mean 2x^2 -3x-1 = 0.  try it
b^2-4ac is called the discriminant
If it is positive, there are two real roots
If it is negative, there are two imaginary roots
If it is zero, there are two real roots with the same value.
For parabolas, x-intercepts occur when y=0
so for 0 = x^2 -5x +6 = (x-2)(x-3), the intercepts are
x=2 and x=3
For x = 5/2, y = -1/4 so the vertex is below the x-axis
For y = -x^2 +2x -1, setting y=0 gives you only x=1 so the vertex is on the x-axis