To determine how many solutions exist for the given system of equations:
[tex]\[
y = x^2 + x + 3
\][/tex]
[tex]\[
y = -2x - 5
\][/tex]
We need to find the points of intersection between the curves of these two equations. Here's a detailed approach to solving this:
1. Set the equations equal to each other:
Since [tex]\( y \)[/tex] is equal in both equations, set the right-hand sides equal:
[tex]\[
x^2 + x + 3 = -2x - 5
\][/tex]
2. Rearrange the equation:
Bring all the terms to one side of the equation to form a standard quadratic equation:
[tex]\[
x^2 + x + 3 + 2x + 5 = 0
\][/tex]
Combine like terms:
[tex]\[
x^2 + 3x + 8 = 0
\][/tex]
3. Solve the quadratic equation:
Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = 8 \)[/tex].
Calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[
\Delta = b^2 - 4ac = 3^2 - 4(1)(8) = 9 - 32 = -23
\][/tex]
4. Analyze the discriminant:
The discriminant [tex]\(\Delta = -23\)[/tex] is less than 0. This indicates that the quadratic equation has no real roots; instead, it has two complex roots.
5. Conclusion:
Since the quadratic equation has no real roots, the system of equations [tex]\( y = x^2 + x + 3 \)[/tex] and [tex]\( y = -2x - 5 \)[/tex] does not have any real points of intersection.
Therefore, the system of equations has:
[tex]\[
\text{No real solutions}
\][/tex]
The correct answer is A. no real solutions
