Question

Use any method to locate the zeros of y = -3x^2 + 12x + 15

Answer 1

To find the zeros of the quadratic equation \( y = -3x^2 + 12x + 15 \), we can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a \), \( b \), and \( c \) are the coefficients of the quadratic equation in the standard form \( ax^2 + bx + c = 0 \).

For our equation:

- \( a = -3 \)
- \( b = 12 \)
- \( c = 15 \)

Let's calculate the discriminant (\( b^2 - 4ac \)) first:

\[
b^2 = 12^2 = 144
\]

\[
4ac = 4 \times (-3) \times 15 = -180
\]

Now, calculate the discriminant:

\[
b^2 - 4ac = 144 - (-180) = 144 + 180 = 324
\]

Since the discriminant is positive, we expect two real roots.

Now let's apply the quadratic formula:

\[
x = \frac{-12 \pm \sqrt{324}}{2 \times -3}
\]

First, we calculate \( \sqrt{324} \):

\[
\sqrt{324} = 18
\]

Now substitute back into the formula:

\[
x = \frac{-12 \pm 18}{-6}
\]

This gives us two cases to solve for \( x \):

1. Using the positive root:

\[
x = \frac{-12 + 18}{-6} = \frac{6}{-6} = -1
\]

2. Using the negative root:

\[
x = \frac{-12 - 18}{-6} = \frac{-30}{-6} = 5
\]

Thus, the zeros of the quadratic equation \( y = -3x^2 + 12x + 15 \) are:

\[
x = -1 \quad \text{and} \quad x = 5
\]