Question

Which quadratic equations have two solutions
y = x^2 -8x -20
y= x^2 + 3
y= -x^2 +2x +1
y = x^2 - 25

Answer 1

y=4x^2 + 12 +9
y = -2x^2 -5x -4

Answer 2

To determine which of the given quadratic equations have two solutions, we can use the discriminant of the quadratic equation. The discriminant \(D\) is given by the formula:

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

For a quadratic equation of the form \(ax^2 + bx + c = 0\):

- If \(D > 0\), there are two distinct real solutions.
- If \(D = 0\), there is exactly one real solution (a repeated solution).
- If \(D < 0\), there are no real solutions.

Now, let's analyze each of the given equations:

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

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

\[
D = (-8)^2 - 4(1)(-20) = 64 + 80 = 144
\]
Since \(D > 0\), this equation has **two solutions**.

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

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

\[
D = 0^2 - 4(1)(3) = 0 - 12 = -12
\]
Since \(D < 0\), this equation has **no real solutions**.

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

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

\[
D = (2)^2 - 4(-1)(1) = 4 + 4 = 8
\]
Since \(D > 0\), this equation has **two solutions**.

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

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

\[
D = 0^2 - 4(1)(-25) = 0 + 100 = 100
\]
Since \(D > 0\), this equation has **two solutions**.

### Summary

The quadratic equations that have two solutions are:

1. \(y = x^2 - 8x - 20\)
2. \(y = -x^2 + 2x + 1\)
3. \(y = x^2 - 25\)

The equation \(y = x^2 + 3\) has no real solutions.

Answer 3

Let's analyze the two quadratic equations given in order to determine how many solutions each has.

1. **Equation: \(y = 4x^2 + 12x + 9\)**

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

Now we calculate the discriminant \(D\):

\[
D = b^2 - 4ac = (12)^2 - 4(4)(9) = 144 - 144 = 0
\]
Since \(D = 0\), this equation has **exactly one solution** (a repeated root).

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

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

Now we calculate the discriminant \(D\):

\[
D = b^2 - 4ac = (-5)^2 - 4(-2)(-4) = 25 - 32 = -7
\]
Since \(D < 0\), this equation has **no real solutions**.

### Summary

- The equation \(y = 4x^2 + 12x + 9\) has **exactly one solution**.
- The equation \(y = -2x^2 - 5x - 4\) has **no real solutions**.