Question

Which quadratic equations have two solutions. Check all that apply
y = x^2 - 8x -20
y = x^2 + 3
y = -x^2 + 2x +1
y = x ^2 - 25
y = 4x^2 + 12x +9
y = -2x ^2 -5x -4

Answer 1

To determine which quadratic equations have two solutions, we can look at the discriminant \( D \) of each quadratic equation of the form \( ax^2 + bx + c = 0 \). The discriminant is given by:

\[
D = b^2 - 4ac
\]

A quadratic equation has:

- Two real distinct solutions if \( D > 0 \)
- One real solution (a double root) if \( D = 0 \)
- No real solutions if \( D < 0 \)

Now let's calculate the discriminant for each of the given quadratic equations:

1. **\( y = x^2 - 8x - 20 \)**

Here, \( a = 1 \), \( b = -8 \), \( c = -20 \).

\[
D = (-8)^2 - 4(1)(-20) = 64 + 80 = 144 \quad (D > 0)
\]

2. **\( y = x^2 + 3 \)**

Here, \( a = 1 \), \( b = 0 \), \( c = 3 \).

\[
D = 0^2 - 4(1)(3) = 0 - 12 = -12 \quad (D < 0)
\]

3. **\( y = -x^2 + 2x + 1 \)**

Here, \( a = -1 \), \( b = 2 \), \( c = 1 \).

\[
D = (2)^2 - 4(-1)(1) = 4 + 4 = 8 \quad (D > 0)
\]

4. **\( y = x^2 - 25 \)**

Here, \( a = 1 \), \( b = 0 \), \( c = -25 \).

\[
D = 0^2 - 4(1)(-25) = 0 + 100 = 100 \quad (D > 0)
\]

5. **\( y = 4x^2 + 12x + 9 \)**

Here, \( a = 4 \), \( b = 12 \), \( c = 9 \).

\[
D = (12)^2 - 4(4)(9) = 144 - 144 = 0 \quad (D = 0)
\]

6. **\( y = -2x^2 - 5x - 4 \)**

Here, \( a = -2 \), \( b = -5 \), \( c = -4 \).

\[
D = (-5)^2 - 4(-2)(-4) = 25 - 32 = -7 \quad (D < 0)
\]

### Summary of Results:

- **Two real solutions:**
- \( y = x^2 - 8x - 20 \) (D = 144)
- \( y = -x^2 + 2x + 1 \) (D = 8)
- \( y = x^2 - 25 \) (D = 100)

- **No real solutions:**
- \( y = x^2 + 3 \) (D = -12)
- \( y = -2x^2 - 5x - 4 \) (D = -7)

- **One real solution:**
- \( y = 4x^2 + 12x + 9 \) (D = 0)

### Conclusion:
The quadratic equations that have **two solutions** are:
- \( y = x^2 - 8x - 20 \)
- \( y = -x^2 + 2x + 1 \)
- \( y = x^2 - 25 \)