Question

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

Answer 1

That is when b^2-4ac>0, that is when you have 2 real solutions

alright, so ax^2+bx+c=0

first one:
a=2, b=-7, c=-9
b^2-4ac=(-7)^2-4(2)(-9)=49+72>0, so this has 2 real solutions

2nd one
a=1, b=-4, c=4
b^2-4ac=(-4)^2-4(1)(4)=16-16=0, so this has only 1 real number solution

3rd
a=4, b=-3, c=-1
b^2-4ac=(-3)^2-4(4)(-1)=9+16>0, so this has 2 real number solutions

4th
a=1, b=-2, c=-8
b^2-4ac=(-2)^2-4(1)(-8)=4+32>0, so this has 2 real number solutions

5th
a=3, b=5, c=3
b^2-4ac=(5)^2-4(3)(3)=25-36<0, this has no real soltuions


answer is 1st, 3rd, 4th
those are
0 = 2x2 – 7x – 9
0 = 4x2 – 3x – 1
0 = x2 – 2x – 8

Answer 2

Answer: $0=2x^2-7x-9\\0=4x^2-3x-1\\0=x^2-2x-8$

Step-by-step explanation:

We know that the standard quadratic equation is  ax^2+bx+c=0

Let's compare all the given equation to it and , find discriminant.

1. a=2, b= -7, c=-9

$b^2-4ac=(-7)^2-4(2)(-9)=49+72>0$

So it has 2 real number solutions.

2. a=1, b=-4, c=4

$b^2-4ac=(-4)^2-4(1)(4)=16-16=0$

So it has only 1 real number solution.

3. a=4, b=-3, c=-1

$b^2-4ac=(-3)^2-4(4)(-1)=9+16=25>0$

So it has 2 real number solutions.

4. a=1, b=-2, c=-8

$b^2-4ac=(-2)^2-4(1)(-8)=4+32=36>0$

So it has 2 real number solutions.

5. a=3, b=5, c=3

$b^2-4ac=(5)^2-4(3)(3)=25-36=-9$

Thus it does not has real solutions.