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
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
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