Question

Problem:

Answer 1

Answer:

Part B. see the procedure

Part C. see the procedure

Step-by-step explanation:

we have

$f(x)=x^{2}+2x+1$ -----> equation A

$g(x)=3-x-x^{2}$ -----> equation B  

Part B. Solve the system algebraically

equate the equation A and the equation B

$x^{2}+2x+1=3-x-x^{2}$

$2x^{2}+3x-2=0$

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$2x^{2}+3x-2=0$

so

$a=2\\b=3\\c=-2$

substitute in the formula

$x=\frac{-3(+/-)\sqrt{3^{2}-4a(2)(-2)}} {2(2)}$

$x=\frac{-3(+/-)\sqrt{25}} {4}$

$x=\frac{-3(+/-)5} {4}$

$x1=\frac{-3(+)5} {4}=0.5$

$x2=\frac{-3(-)5} {4}=-2$

Find the values of y

For x=0.5

$f(0.5)=0.5^{2}+2(0.5)+1=2.25$

For x=-2

$f(-2)=(-2)^{2}+2(-2)+1=1$

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