To solve the system of equations given by [tex]\( y = 3x^2 + 18x + 24 \)[/tex] and [tex]\( y = 2x + 5 \)[/tex], we need to find the points where these two functions intersect. We can do this by setting the equations equal to each other and solving for [tex]\( x \)[/tex].
1. Start by setting the equations equal to each other:
[tex]\[
3x^2 + 18x + 24 = 2x + 5
\][/tex]
2. Move all terms to one side of the equation to set it to zero:
[tex]\[
3x^2 + 18x + 24 - 2x - 5 = 0
\][/tex]
[tex]\[
3x^2 + 16x + 19 = 0
\][/tex]
3. Solve the quadratic equation [tex]\( 3x^2 + 16x + 19 = 0 \)[/tex]. This is a standard quadratic equation and can be solved using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\( a = 3 \)[/tex], [tex]\( b = 16 \)[/tex], and [tex]\( c = 19 \)[/tex].
4. Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[
x = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 3 \cdot 19}}{2 \cdot 3}
\][/tex]
[tex]\[
x = \frac{-16 \pm \sqrt{256 - 228}}{6}
\][/tex]
[tex]\[
x = \frac{-16 \pm \sqrt{28}}{6}
\][/tex]
[tex]\[
x = \frac{-16 \pm 2\sqrt{7}}{6}
\][/tex]
Simplify the fraction:
[tex]\[
x = \frac{-8 \pm \sqrt{7}}{3}
\][/tex]
5. Approximate the solutions, rounding to the nearest hundredth:
[tex]\[
x_1 \approx -3.55, \quad x_2 \approx -1.78
\][/tex]
6. Find the corresponding [tex]\( y \)[/tex]-values for each [tex]\( x \)[/tex]-value using the linear equation [tex]\( y = 2x + 5 \)[/tex]:
[tex]\[
y_1 = 2(-3.55) + 5 \approx -2.10
\][/tex]
[tex]\[
y_2 = 2(-1.78) + 5 \approx 1.43
\][/tex]
7. The points of intersection are:
[tex]\[
(x_1, y_1) = (-3.55, -2.10)
\][/tex]
[tex]\[
(x_2, y_2) = (-1.78, 1.43)
\][/tex]
Therefore, the solutions are:
[tex]\[
(x, y) = (-3.55, -2.10) \quad \text{or} \quad (x, y) = (-1.78, 1.43)
\][/tex]
