SOLUTION: Substitute the values for a, b, and c into b2 – 4ac to determine the discriminant. Which quadratic equations will have two real number solutions? (The related quadratic function wi

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Question 1041402: Substitute the values for a, b, and c into b2 – 4ac to determine the discriminant. Which quadratic equations will have two real number solutions? (The related quadratic function will have two x-intercepts.) Check all that apply.
0 = 2x2 – 7x – 9
0 = x2 – 4x + 4
0 = 4x2 – 3x – 1
0 = x2 – 2x – 8
0 = 3x2 + 5x + 3
Found 2 solutions by Boreal, Edwin McCravy:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
2x^2-7x-9
a=2;b=-7;c=-9. b^2-4ac=49+18=67. This will have two real intercepts

x^-4x+4
a=1;b=-4;c=4. b^2-4ac=0
This is one root of multiplicity 2. The factors are (x-2)^2, and the graph will "bounce" at the root of x=2

4x^2-3x-1
a=4;b=-3;c=-1 b^2-4ac=9+16=25. Two real roots

x^2-2x-8
a=1;b=-2;c=-8 b^2-4ac=4+31=36. Two real roots.

3x^2+5x+3
a=3;b=5;c=3 b^2-4ac=25-35=-11. No real roots
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
He told you correctly.  But here's the rule
that you go by:

For equations of this type:

0 = ax2 + bx + c

a = coefficient of x2.
b = coefficient of x.
c = the constant term.

Calculate b2-4ac.

If you get a positive number, there are two real solutions.
 (The related quadratic function will have two x-intercepts)

If you get 0, there is one real solution.
  (The related quadratic function will just touch the x-axis in one point.)

If you get a negative number, there are no real solutions.
 (The related quadratic function will not touch or cross the x-axis)

Edwin
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