Question

Can someone please check these?
Discriminants and real number solutions

Answer 1

1.
a = 3, b = 2, c = 1
Formula = b^2 - 4ac
2^2 - 4 * 3 * 1 =
4 - 4 * 3 * 1 =
4 - 12 * 1 =
4 - 12 =
-8
Discriminant = -8
The quadratic equation has no real number solution, because the discriminant is a negative number.

2.
a = 1, b = 4, c = 4
Formula = b^2 - 4ac
4^2 - 4 * 1 * 4
16 - 4 * 1 * 4 =
16 - 4 * 4 =
16 - 16 =
0
Discriminant = 0
The quadratic equation has one real solution because the discriminant is 0.

3.
5x^2 + x - 4 = 4 - 4
5x^2 + x - 4 = 0
a = 5, b = 1, c = -4
Formula = b^2 - 4ac
1^2 - 4 * 5 * (-4) =
1 - 4 * 5 * (-4) =
1 - 20 * (-4) =
1 - (-80) =
81
Discriminant = 81
The quadratic equation has two real number solutions because the discriminant is a positive number.

4.
2x^2 - 3x + 5 = -5 + 5
2x^2 - 3x + 5 = 0
a = 2, b = -3, c = 5
Formula = b^2 - 4ac
-3^2 - 4 * 2 * 5 =
9 - 4 * 2 * 5 =
9 - 8 * 5 =
9 - 40 =
-31
Discriminant = -31
The quadratic equation has no real number solutions, because the discriminant is a negative number.

5.
x^2 - x - 12 = 12 - 12
x^2 - x - 12 = 0
a = 1, b = -1, c = -12
-1^2 - 4 * 1 * (-12) =
1 - 4 * 1 * (-12) =
1 - 4 * (-12) =
1 - (-48) =
49
Discriminant = 49
The quadratic equation has two real number solutions, because the discriminant is a positive number.

6.
x^2 - 2x + 1 = -1 + 1
x^2 - 2x + 1 = 0
a = 1, b = -2, c = 1
Formula = b^2 - 4ac
-2^2 - 4 * 1 * 1 =
4 - 4 * 1 * 1 =
4 - 4 * 1 =
4 - 4 =
0
Discriminant = 0
The quadratic equation has one real number solution because the discriminant is 0.

Answer 2

@iambatman

Answer 3

I looked at the first three and the process you're using was correct for each of them. You more than likely didn't make any mistakes :)

Answer 4

Thanks:)

Answer 5

You're welcome