Chapter Closure
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Exercise 114 Page 206
a
Graph the circle and the parabola on one coordinate plane.
b
Use the Substitution Method, then recall the Quadratic Formula.
a
Solutions: Approximately (1.35,4.82) and (-1.35,4.82)
Graph:
b
x=sqrt(-1±sqrt(113)/2-3)≈ ± 1.35 and y=-1+sqrt(113)/2≈ 4.82
a
To solve the system of equations by graphing, we will draw the circle and the graph of the quadratic function on the same coordinate grid. Let's start with the circle!
Graphing the Circle
We are given an equation in the standard form of a circle.
x^2+y^2= r^2
Recall that this type of equation describes a circle with the center (0,0) and a radius r. Looking at our equation, we can see that the radius is 5.
Graphing the Parabola
We want to draw the graph of the given equation. It has the form y=x^2+ k, where k is either a positive or a negative number. y=x^2+ 3 To do so, we will first draw the graph of its parent quadratic function, y=x^2. Recall that the graph of y=x^2 is a parabola opening upwards, with vertex (0,0). Its axis of symmetry is the vertical line x=0, and it passes through the points (1,1) and (- 1,1).
Our function is a vertical translation of the parent function by 3 units in the positive direction. Thus, we will translate the above graph 3 units upwards
Finding the Solutions
Finally, let's try to identify the coordinates of the points of intersection of the circle and the parabola.
It looks like the points of intersection occur at approximately (1.4,4.8) and (-1.4,4.8).
b
Since y is isolated in the second equation, we will solve the given system of equations using the Substitution Method.
x^2+y^2=25 & (I) y=x^2+3 & (II) We can start with substituting the value of y into the first equation.
x^2+y^2=25 y=x^2+3
Substitute
(I): y= x^2+3
x^2+( x^2+3)^2=25 y=x^2+3
▼
(I): Simplify
ExpandNegPerfectSquare
(I):(a-b)^2=a^2-2ab+b^2
x^2+x^4+6x^2+9=25 y=x^2+3
AddTerms
(I): Add terms
x^4+7x^2+9=25 y=x^2+3
SubEqn
(I):LHS-25=RHS-25
x^4+7x^2-16=0 y=x^2+3
The next step will be to find the solutions of the first equation. To make this a little bit simpler, we can set a new variable t that is equal to x^2 and rewrite Equation (I) using only t-terms. x^4+7x^2-16=0 ⇒ t^2+7t-16=0
t=- b±sqrt(b^2-4ac)/2a
SubstituteValues
Substitute values
t=- 7±sqrt(( 7)^2-4( 1)( -16))/2( 1)
▼
Solve for t
CalcPow
Calculate power
t=-7±sqrt(49-4(1)(-16))/2(1)
MultByOne
a * 1=a
t=-7±sqrt(49-4(-16))/2
MultNegNegOnePar
- a(- b)=a* b
t=-7±sqrt(49+64)/2
AddTerms
Add terms
t=-7±sqrt(113)/2
▼
Simplify
Rewrite
Rewrite -7 as -1-6
t=-1-6±sqrt(113)/2
CommutativePropMult
Commutative Property of Multiplication
t=-1±sqrt(113)-6/2
WriteDiffFrac
Write as a difference of fractions
t=-1±sqrt(113)/2-6/2
CalcQuot
Calculate quotient
t=-1±sqrt(113)/2-3
This result tells us that we have two solutions for t. One of them will use the positive sign, and the other one will use the negative sign. Since t is equal to the squared value, it can be only a positive number. Therefore, our only possible solution for t is -1+sqrt(113)2-3. Using this value, we are able to find the values of x.
t=x^2
Substitute
t= -1+sqrt(113)/2-3
-1+sqrt(113)/2-3=x^2
RearrangeEqn
Rearrange equation
x^2=-1+sqrt(113)/2-3
SqrtEqn
sqrt(LHS)=sqrt(RHS)
x=±sqrt(-1+sqrt(113)/2-3)
Now, consider Equation (II). y=x^2+3 We can substitute the values of x into the above equation to find the values for y. Let's start with the positive one.
y=x^2+3
Substitute
x= sqrt(-1+sqrt(113)/2-3)
y=( sqrt(-1+sqrt(113)/2-3))^2+3
PowSqrt
( sqrt(a) )^2 = a
y=-1+sqrt(113)/2-3+3
AddTerms
Add terms
y=-1+sqrt(113)/2
To find the other solution, we will substitute -sqrt(-1+sqrt(113)2-3) for x in Equation (II) again.
y=x^2+3
Substitute
x= -sqrt(-1+sqrt(113)/2-3)
y=( -sqrt(-1+sqrt(113)/2-3))^2+3
NegBaseToPosBase
(- a)^2=a^2
y=(sqrt(-1+sqrt(113)/2-3))^2+3
PowSqrt
( sqrt(a) )^2 = a
y=-1+sqrt(113)/2-3+3
AddTerms
Add terms
y=-1+sqrt(113)/2
We found that the value of y is -1+sqrt(113)2 for x=±sqrt(-1+sqrt(113)2-3). We can approximate our solution using a calculator. x=±sqrt(-1+sqrt(113)/2-3)≈ ± 1.35 y=-1+sqrt(113)/2 ≈ 4.82
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