Answer :
Answer:
Part B. see the procedure
Part C. see the procedure
Step-by-step explanation:
we have
[tex]f(x)=x^{2}+2x+1[/tex] -----> equation A
[tex]g(x)=3-x-x^{2}[/tex] -----> equation B
Part B. Solve the system algebraically
equate the equation A and the equation B
[tex]x^{2}+2x+1=3-x-x^{2}[/tex]
[tex]2x^{2}+3x-2=0[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]2x^{2}+3x-2=0[/tex]
so
[tex]a=2\\b=3\\c=-2[/tex]
substitute in the formula
[tex]x=\frac{-3(+/-)\sqrt{3^{2}-4a(2)(-2)}} {2(2)}[/tex]
[tex]x=\frac{-3(+/-)\sqrt{25}} {4}[/tex]
[tex]x=\frac{-3(+/-)5} {4}[/tex]
[tex]x1=\frac{-3(+)5} {4}=0.5[/tex]
[tex]x2=\frac{-3(-)5} {4}=-2[/tex]
Find the values of y
For x=0.5
[tex]f(0.5)=0.5^{2}+2(0.5)+1=2.25[/tex]
For x=-2
[tex]f(-2)=(-2)^{2}+2(-2)+1=1[/tex]
the solutions are the points
(0.5,2.25) and (-2,1)
Part C. Solve the system by graph
using a graphing tool
we know that
The solution of the non linear system is the intersection point both graphs
The intersection points are (0.5,2.25) and (-2,1)
therefore
The solutions are the points (0.5,2.25) and (-2,1)
see the attached figure
