Exercise 52 Page 267

For simplicity, we will solve the given system of equations using the Substitution Method. x^2+y^2=4 & (I) y^2=x+2 & (II) The y^2-variable is isolated in Equation (II). This allows us to substitute its value x+2 for y^2 in Equation (I). Notice that in Equation (I) we have a quadratic equation in terms of only the x-variable. Notice that there are many ways to solve the obtained equation. We will use the Quadratic Formula. Let's determine a, b, and c. x^2+x-2=0 ⇔ 1x^2+ 1x+( - 2)=0We see above that a= 1, b= 1, and c= - 2. Let's substitute these values into the Quadratic Formula to solve the equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 1±sqrt(1^2-4( 1)( - 2))/2( 1)

▼

Simplify right-hand side

CalcPow

Calculate power

x=- 1±sqrt(1-4(1)(- 2))/2(1)

Multiply

Multiply

x=- 1±sqrt(1+8)/2

SubTerms

Subtract terms

x=- 1±sqrt(9)/2

CalcRoot

Calculate root

x=- 1± 3/2

We can calculate the first root using the positive sign and the second root using the negative sign.

x=- 1± 3/2
x=- 1+ 3/2	x=- 1- 3/2
x=1	x=- 2

Now consider Equation (II). y^2=x+2 We can substitute x=1 and x=- 2 into the above equation to find the values for y. Let's start with x=1.

y^2=x+2

Substitute

x= 1

y^2= 1+2

AddTerms

Add terms

y^2=3

SqrtEqn

sqrt(LHS)=sqrt(RHS)

y=±sqrt(3)

We found that y=±sqrt(3), when x=1. This means we get two solutions of the system: (1,sqrt(3)) and (1,-sqrt(3)). To find the other solution, we will substitute - 2 for x in Equation (II) again.

y^2=x+2

Substitute

x= - 2

y^2= - 2+2

AddTerms

Add terms

y^2=0

SqrtEqn

sqrt(LHS)=sqrt(RHS)

y=0

We found that y=0, when x=- 2. Therefore, our third solution is (- 2,0).

Graphing

Notice that Equation (I) represents a circle and Equation (II) represents a parabola. If you need explanations on graphing circles, please refer to this example. If you need explanations on graphing parabolas, please refer to this other example. Let's graph them and plot the points (1,sqrt(3)), (1,-sqrt(3)), and (- 2,0).

Both the circle and the parabola intersect at the points (1,sqrt(3)), (1,-sqrt(3)), and (- 2,0), which confirms our answer.