Question

Y equals x squared + 3x + 8 what is the X intercepts

Answer 1

To find the x-intercepts of the quadratic equation \( y = x^2 + 3x + 8 \), we need to set \( y \) to 0 and solve for \( x \):

\[
0 = x^2 + 3x + 8
\]

Now, we can rearrange it to:

\[
x^2 + 3x + 8 = 0
\]

Next, we'll use the quadratic formula to solve for \( x \):

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

In this equation, \( a = 1 \), \( b = 3 \), and \( c = 8 \).

First, we calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = 3^2 - 4 \times 1 \times 8 = 9 - 32 = -23
\]

Since the discriminant is negative (\(-23\)), this means there are no real x-intercepts. The quadratic does not cross the x-axis.

Thus, the conclusion is that the equation \( y = x^2 + 3x + 8 \) has no real x-intercepts.