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How many solutions does this system of equations have?

[tex]\[
\begin{array}{l}
y = 2x + 5 \\
y = x^3 + 4x^2 + x + 2
\end{array}
\][/tex]

A. No solution
B. 1 solution
C. 2 solutions
D. 3 solutions

Sagot :

To determine how many solutions the system of equations has, let's analyze the equations given:

[tex]\[ \begin{array}{l} y = 2x + 5 \\ y = x^3 + 4x^2 + x + 2 \end{array} \][/tex]

We need to find the points [tex]\((x, y)\)[/tex] where both equations are equal to each other. To do this, we set the right-hand sides of the equations equal to each other:

[tex]\[ 2x + 5 = x^3 + 4x^2 + x + 2 \][/tex]

First, move all terms to one side of the equation to set it to zero:

[tex]\[ 0 = x^3 + 4x^2 + x + 2 - 2x - 5 \][/tex]

Simplify by combining like terms:

[tex]\[ 0 = x^3 + 4x^2 - x - 3 \][/tex]

Now, we need to solve the cubic equation:

[tex]\[ x^3 + 4x^2 - x - 3 = 0 \][/tex]

A cubic equation can have up to three real solutions. By solving this equation, we find the values of [tex]\(x\)[/tex] that satisfy it.

The result shows that the number of solutions to this equation is three, and there are three corresponding [tex]\(x\)[/tex] values:

[tex]\[ \left(-\frac{4}{3} - \frac{19}{3(a)} - \frac{a}{3}, -\frac{4}{3} - \frac{a}{3} - \frac{19}{3(b)}, -\frac{4}{3} - \frac{b}{3} - \frac{19}{3c}\right) \][/tex]

where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are complex expressions involving cube roots and imaginary numbers.

Since we are only asked for the number of solutions, we can conclude that there are three solutions to the system of equations.

Thus, the correct answer is:

[tex]\[ \boxed{D. 3 solutions} \][/tex]