Hello there. To solve this question, we'll have to remember some properties about system of equations.
Given the following system of equations:
[tex]\begin{cases}y=x^2+2x+2 \\ y=x+1\end{cases}[/tex]In the R² plane, we have that the solutions to this system are the points of intersection of the parabola y = x² + 2x + 2 and the line y = x + 1.
In fact, there must be two points of intersection and, therefore, two solutions for this system of equation.
Let's plug the first equation into the second equation (called the method of substitution):
[tex]x^2+2x+2=x+1[/tex]Subtract x + 1 on both sides of the equation
[tex]\begin{gathered} x^2+2x+2-(x+1)=x+1-(x+1) \\ x^2+2x+2-x-1=0 \\ x^2+x+1=0 \end{gathered}[/tex]Now, this is a quadratic equation of the form:
[tex]ax^2+bx+c=0,a\ne0[/tex]Its solutions are given by the quadratic formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Plugging the coefficients a = b = c = 1, we get:
[tex]x=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot1}}{2\cdot1}[/tex]Square the number and multiply the values. Add everything inside the radical.
[tex]x=\frac{-1\pm\sqrt[]{1-4}_{}}{2}=\frac{-1\pm\sqrt[]{-3}_{}}{2}[/tex]In this case, we got a negative number inside the square root. This is not defined in the real numbers, therefore we don't have solutions for this system of equations.
Graphically, and this is why I omitted it from the beginning, is that in fact the parabola and the line won't have any intersection points, as you can see in the following image:
If we were to solve this system of equations in the set of the complex numbers, then we would have solutions.
First, remember:
[tex]\sqrt[]{-1}=i[/tex]Such that:
[tex]x=\frac{-1\pm\sqrt[]{-3}}{2}=\frac{-1\pm\sqrt[]{3}\cdot\sqrt[]{-1}}{2}=\frac{-1\pm\sqrt[]{3}i}{2}[/tex]And the values of y can be found by plugging it in the second equation (the easier one)
[tex]y=\frac{-1\pm\sqrt[]{3}i}{2}+1=\frac{-1\pm\sqrt[]{3}i+2_{}}{2}=\frac{1\pm\sqrt[]{3}i}{2}[/tex]And the ordered pairs (x, y) would have been:
[tex]\begin{gathered} \mleft(\dfrac{-1+\sqrt{3}i}{2},\dfrac{1+\sqrt{3}i}{2}\mright) \\ \mleft(\dfrac{-1-\sqrt{3}i}{2},\dfrac{1-\sqrt{3}i}{2}\mright) \end{gathered}[/tex]But as we're talking about real numbers, we say that this is an impossible system of equations, that has no solutions.
