Solving Systems of Equations Including Quadratics

a= 2, b= -8

x = - -8/2( 2)

▼

Solve for x

x = - -8/4

Calculate quotient

x = -(-2)

- (- a)=a

x = 2

This indicates that the axis of symmetry is a vertical line on x=2.

Then, to find the vertex of the parabola, the equation is evaluated to find the value of y when x=2.

2x^2 - 8x + 6 = y

x= 2

2( 2)^2 - 8( 2) + 6 = y

▼

Solve for y

Calculate power

2(4) - 8(2) + 6 = y

8 - 16 + 6 = y

Add terms

-2 = y

Rearrange equation

y = - 2

This means that the vertex lies on point (2,-2).

Then, the y-intercept is the value of c of the equation, which in this case is 6. Since the axis of symmetry divides the graph into two mirror images, the parabola also goes through point in (4,6).

Using these points, it is possible to draw the parabola.

Finding the Points of Intersection

Finally, finding the points in which the graph intersect, the solutions of the nonlinear system are found.

This means that the points (1,0) and (4,6) are the solutions of the nonlinear system.

Example

Graphing a Linear-Quadratic System

Emily received a scholarship this summer and attended a camp for skydiving. She has always dreamed of feeling like flying in the sky. Surprisingly, her first task given by her instructor Sky Flyer is to solve a nonlinear system graphically.

Instructor Sky Flyer must be up to something by having Emily solve this system. 2x - y = 0 & (I) x^2 + 2x - 1 = y & (II) Along with Emily, solve the nonlinear system graphically.

Answer

Hint

Graph both equations and find the points of intersection.

Solution

To solve this system graphically, the points of intersection of the graphs need to be found. To do so, each equation should be graphed. Equation (I) is a linear equation that can be written in slope-intercept form. 2x - y =0 ⇒ y = 2x The slope of 2 and the y-intercept of 0 can be used to graph this line in the coordinate plane.

Equation (II) is a quadratic equation written in standard form. ax^2 + bx + c = y ⇓ 1x^2 + 2x + ( - 1) = y The first step to graph this equation is to find the axis of symmetry. That can be done by substituting the values of a = 1 and b = 2 into the Quadratic Formula to find the axis. Note that the root term is 0.

x = - b± 0/2a

IdPropAdd

Identity Property of Addition

x = - b/2a

a= 1, b= 2

x = - 2/2( 1)

x = - 2/2

Calculate quotient

x = -1

This indicates that the axis of symmetry is the vertical line x=-1.

Then, to find the vertex of the parabola, -1 will be substituted for x in Equation (II).

x^2 + 2x - 1 = y

x= -1

( -1)^2 + 2( -1) - 1 = y

▼

Simplify

Calculate power

1 + 2(-1) -1 = y

1 - 2 - 1 = y

Subtract terms

-2 = y

Rearrange equation

y = -2

This indicates that the vertex is at point (-1, -2).

The value of c on a quadratic equation written in standard form indicates the value of the y-intercept. In this case, the value of the y-intercept is -1. Since the axis of symmetry divides the graph into mirror images, the points (0,-1) and (-2,-1) can be added.

These points can be used to graph the parabola.

Looking at the graph, the points of intersection can be located.

The solutions of the nonlinear system are the points (-1,-2) and (1,2).

Example

Graphing a Quadratic-Quadratic System

Instructor Sky Flyer continues teaching Emily about nonlinear systems. Emily is unsure but thinks they must be getting closer to applying this knowledge to something spectacular.

Sky Flyer gives the following quadratic-quadratic system to Emily and the other students learning to skydive. y = 16x^2 + 2x + 5 & (I) y = - 12x - 2x +5 & (II) Along with Emily, solve the nonlinear system graphically.

Answer

Hint

How do you find the axis of symmetry of a parabola if the quadratic equation is written in standard form?

Solution

To solve a quadratic-quadratic system graphically, both quadratic equations are graphed to identify the points of intersection. Looking at the system, it can be noted that both equations of the system are written in standard form.

y= ax^2 + bx + c
y = 1/6x^2 + 2x + 5	y = -1/2x + ( -2)x + 5

The graphs of both quadratic equations are parabolas. The values of a and b can be used to find the axis of symmetry of each graph. To do so, the values are substituted into a formula. This can be done for Equation (I).

x = - b/2a

a= 16, b= 2

x = - 2/2( 16)

▼

Solve for x

CancelCommonFac

Cancel out common factors

x = - 2/2( 16)

SimpQuot

Simplify quotient

x = - 1/16

DivOneByFracD

1/a/b= b/a

x = - 6

This indicates that the axis of symmetry of the parabola of Equation (I) is the vertical line x=-6.

To find the vertex of the parabola, the value x=-6 is substituted into Equation (I).

y = 1/6x^2 + 2x + 5

x= - 6

y = 1/6( -6)^2 + 2( -6) + 5

▼

Solve for y

Calculate power

y = 1/6(36) + 2(-6) + 5

y = 1/6(36) - 12 + 5

MoveRightFacToNum

a/c* b = a* b/c

y = 36/6 - 12 + 5

Calculate quotient

y = 6 - 12 + 5

Add terms

y = -1

This indicates that the vertex of the first parabola lies in the point (-6,-1).

It is possible to find the y-intercept of the parabola with the value of c. In Equation (I), this value is 5. Also, since the axis of symmetry mirrors the image of the parabola, there is another known point at (-12, 5). These points can be used to graph the parabola.

The same can be done to graph Equation (II). The axis of symmetry can be found by substituting the values of a and b in the formula.

x = - b/2a

a= - 12, b= -2

x = - -2/2( - 12)

▼

Solve for x

MultPosNeg

a(- b)=- a * b

x =- -2/-2( 12)

MultFracByInverse

a/b* b/a=1

x = --2/-1

Calculate quotient

x = -2

This indicates that the axis of symmetry is the vertical line at x=-2.

Also, to find the vertex of the second parabola, x=-2 is substituted into Equation (II).

y = -1/2x^2 - 2x + 5

x= - 2

y = -1/2( -2)^2 - 2( -2) + 5

▼

Solve for y

Calculate power

y = - 1/2(4) - 2(-2) + 5

Calculate quotient

y = - 2 - 2(-2) + 5

y = - 2 + 4 + 5

Add terms

y = 7

Now it is known that the vertex of the second parabola lies at (-2,7).

The value c=5 on Equation (II) indicates that the y-intercept of the second parabola is y=5. Since the axis of symmetry mirrors the image, there is another known point (-4,5).

Now that the two equations are graphed, the solutions of the nonlinear system can be determined. They are the points of intersection of the graphs.

The points (0,5) and (-6,-1) are the solutions of the nonlinear system.

Discussion

Solving Nonlinear Systems Using Elimination

Just like in a system of linear equations, there are many methods that can be used to solve a nonlinear system.

Some nonlinear systems can be solved by eliminating one of the variables of the equations involved, similar to the elimination method used in systems of linear equations. Consider the following example. - 2x^2 + 8x + 16 = 2y & (I) x+4 = y & (II) Since Equation (I) is a quadratic equation, the system is nonlinear. The system can be solved by elimination following the steps below.

Find a Variable to Eliminate

There are two different variables in the nonlinear system, x and y. The equations have different powers for the terms of x, as one is quadratic and the other is linear. An the other hand, both equations have linear terms of y. Therefore, variable y will be eliminated. - 2x^2 + 8x + 16 = 2y & (I) x+4 = y & (II)

Eliminating the Variable

For a variable to be eliminated, its coefficient has to be the same in both equations. This can be achieved by dividing Equation (I) by 2 and then subtracting Equation (II) from Equation (I).

- 2x^2 + 8x + 16 = 2y x+4 = y

DivEqn

(I): .LHS /2.=.RHS /2.

- x^2 + 4x + 8 = y x+4 = y

(I): Subtract (II)

- x^2 + 4x + 8 - ( x + 4)= y - y x+4 = y

(I): Subtract terms

- x^2 + 3x + 4 = 0 x+4 = y

It can be noted that one of the equations only depends on x now.

Solving the Single Variable Equation

Equation (II) is a quadratic equation in terms of x. The equation is already written in standard form. ax^2 + bx + c = 0 ⇓ ( -1 )x^2 + ( 3)x + 4 = 0 This equation can be solved using the quadratic formula.

- x^2 + 3x + 4 = 0

UseQuadForm

Use the Quadratic Formula: a = -1, b= 3, c= 4

x=- 3±sqrt(3^2 - 4( -1)( 4))/2( -1)

Calculate power

x=-3±sqrt(9 - 4(-1)(4))/2(-1)

x=-3±sqrt(9 + 16)/-2

Add terms

x=-3±sqrt(25)/-2

Calculate root

x = -3 ± 5/-2

There are two possible values for x. These can be found by adding or subtracting 5.

x = -3 ± 5/-2
x = -3 + 5/-2	x = -3 - 5/-2
x = 2/-2	x = -8/-2
x = -1	x = 4

Therefore, the values of x that solve the nonlinear system are -1 and 4.

Substituting the Values

Now that there are two known values for x, the solution values for y can be found by substituting the values of x into either equation. Since Equation (II) is linear, it is easier to do so in this equation.

x + 4 = y
-1 + 4 = y	4+4 = y
y = 3	y = 8

Combining each x-value value with its correspondent y-value, it can be seen that the solutions can be written as the points (-1,3) and (4,8).

It should be noted that not every nonlinear system can be solved using this method.

Example

Skydiving

Instructor Sky Flyer takes the students out to the yard. Emily, super excited, looks up and sees someone free falling with a parachute!

The height above the ground of the skydiver can be modeled by a quadratic equation. h = -4.9t^2 + 5450 In this equation, t is the time passed in seconds after the jump and h is the skydiver's height above the ground in meters. After releasing her parachute, the rate at which she is falling slows. The model for that height is given by the following linear equation. h = -5t + 4450 How long after jumping did the jumper release their parachute? Use the elimination method to find the solution, and round that time to one decimal place.

Hint

Which variable is the easiest to eliminate?

Solution

First of all, the given equations can be combined to make a nonlinear system. h = -4.9t^2 + 5450 & (I) h = -5t + 4450 & (II) Finding how long after jumping did the jumper release his parachute is the same as finding the time t where the model changed from Equation (I) to Equation (II). This can be done by solving the nonlinear system. The first step to solve the nonlinear system by elimination is to find a variable to eliminate. h = -4.9t^2 + 5450 h = -5t + 4450 The variable h is chosen because there are only linear terms of that variable. Then, the elimination is done by subtracting Equation (II) from Equation (I).

h = -4.9t^2 + 5450 h = -5t + 4450

(I): Subtract (II)

h - h = -4.9t^2 + 5450 - ( -5t + 4450) h = -5t + 4450

(I): Subtract terms

0 = - 4.9 t^2 + 5t + 1000 h = -13t + 450

Now a quadratic equation was obtained. The equation is already written in standard form. at^2 + bt + c = 0 ⇓ -4.9t^2 + 5t + 1000 = 0 This equation is solved using the quadratic formula.

t=- b±sqrt(b^2 - 4ac)/2a

Substitute values

t=- ( 5)±sqrt(5^2-4( - 4.9)( 1000))/2( -4.9)

Calculate power

t=-5±sqrt(25 - 4(-4.9)(1000))/2(-4.9)

t=-5±sqrt(25 + 19 600)/-9.8

Add terms

t=-5±sqrt(19 625)/-9.8

The ± sign indicates that there are two possible values for t. These can be found by either adding or subtracting the value of the root.

t=-5±sqrt(19 625)/-9.8
t = -5 + sqrt(19 625)/-9.8	t = -5 - sqrt(19 625)/-9.8
t = - 13.7846	t = 14.8050
t ≈ -13.8	t ≈ 14.8

The two possible values for t are about 14.8 seconds and about -13.8 seconds. Only positive numbers make sense in this case since the skydiver cannot wait a negative time before releasing the parachute. This means that the jumper waited about 14.8 seconds to release the parachute.

Example

Acrobats on the Circus

Before having the chance to skydive, Instructor Sky Flyer wants to show how the students can train by practicing acrobatics. They all walk into a large gym to see a few pros practice. One acrobat jumped from a platform, and another dove from another higher platform just a split second later. Mid-air, they did a high-five while spinning!

The heights above the floor of both acrobats at which they did a high five can be modeled using a system of quadratic equations. These heights are measured from the moment the first acrobat jumps. h = -16(t-1)^2+16 & (I) h = -16(t-2)^2+16 & (II) In these equations, h is the distance of the acrobats from the floor in feet, and t is the time in seconds.

a How long after jumping from their platforms did the acrobats high-five? Solve using elimination.

b At what height above the floor were the acrobats when they did the high-five?

Hint

a Which variable is present in the least amount of terms?

b Substitute the value from Part A into either equation.

Solution

a The first step to solve the given system using elimination is to find a variable to eliminate. The objective is to obtain a single equation that depends on a single variable. The variable h is present in only one term, so it can be eliminated to obtain an equation in terms of t.

h = -16(t-1)^2+16 & (I) h = -16(t-2)^2+16 & (II) To eliminate the variable, Equation (II) is subtracted from Equation (I). To get started, it is practical to expand the perfect squares inside each parenthesis.

h = -16(t-1)^2+16 h = -16(t-2)^2+16

(I): (a-b)^2=a^2-2ab+b^2

h = -16(t^2-2t+1)+16 h = -16(t-2)^2+16

(II): (a-b)^2=a^2-2ab+b^2

h = -16(t^2-2t+1)+16 h = -16(t^2-4t+4)+16

Distr

(I): Distribute - 16

h = -16 t^2 + 32t - 16+16 h = -16(t^2-4t+4)+16

Distr

(II): Distribute -16

h = -16 t^2 + 32t - 16+16 h = -16 t^2 + 64t - 64+16

Add terms

h = -16 t^2 + 32t h = -16 t^2 + 64t - 48

(I): Subtract (II)

h - h = -16 t^2 + 32t - ( -16 t^2 + 64t - 48) h = -16 t^2 + 64t - 48

(I): Subtract terms

0 = -32t + 48 h = -16 t^2 + 64t - 48

A linear equation that depends on t was obtained. Solving this equation gives the time at which the acrobats did the high-five.

0 = -32t + 48

LHS--32t=RHS--32t

32t = 48

DivEqn

.LHS /32.=.RHS /32.

t = 48/32

ReduceFrac

a/b=.a /16./.b /16.

t = 3/2

This means that the acrobats did the high-five 32, or 1.5 seconds after the first one jumped.

b To find the height above the floor at which the acrobats did the high-five, the time found in Part A needs to be substituted into either equation of the given system. In this case, Equation (I) will be used.

h = -16(t-1)^2+16

t= 1.5

h = -16( 1.5-1)^2+16

SubTerm

Subtract term

h = -16(0.5)^2+16

Calculate power

h = -16(0.25)+16

h = -4+16

Add terms

h = 12

The height at which the acrobats did the high-five is 12 feet.

Discussion

Solving Nonlinear Systems Using Substitution

In a similar way that a system of linear equations can be solved using the substitution method, there are nonlinear systems that can be solved by substituting. As an example, consider the following linear-quadratic system. y = x^2 + 2x - 19 & (I) y = 5x - 1 & (II) This system is solved using substitution following the steps below.

Finding a Term to Substitute

The first step to solve a nonlinear system using substitution is to identify which term is substituted from one equation to the other. The given system has the y variable already isolated so it is easy to select that term. y = x^2 + 2x - 19 & (I) y = 5x - 1 & (II)

Substituting a Term

Now the value of y from Equation (II) is substituted into Equation (I). Then, the resulting equation is simplified as much as possible.

y = x^2 + 2x - 19 y = 5x - 1

(I): y= 5x - 1

5x - 1 = x^2 + 2x - 19 y = 5x - 1

(I): LHS-5x=RHS-5x

- 1 = x^2 - 3x - 19 y = 5x - 1

AddEqn

(I): LHS+1=RHS+1

0 = x^2 - 3x - 18 y = 5x - 1

(I): Rearrange equation

x^2 - 3x - 18 = 0 y = 5x - 1

A quadratic equation that only depends on x was obtained.

Solving the Resulting Equation

A quadratic equation was obtained from the previous step. This equation is written in standard form. ax^2 + bx + c = 0 ⇓ 1x^2 + ( -3)x + ( -18) = 0 These values can be substituted into the quadratic formula.

x=- b±sqrt(b^2 - 4ac)/2a

Substitute values

x=-( -3)±sqrt(( - 3)^2 - 4( 1)( -18 ))/2( 1)

Calculate power

x=-(-3)±sqrt(9 - 4(1)(-18))/2(1)

x=-(-3)±sqrt(9 +72)/2

- (- a)=a

x=3±sqrt(9 +72)/2

Add terms

x=3±sqrt(81)/2

Calculate root

x = 3 ± 9/2

There are two possible values for x depending if there is a subtraction or an addition. These values are calculated individually.

x=3± 9/2
x_1 = 3+9/2	x_2 = 3-9/2
x_1 = 12/2	x_2 = -6/2
x_1 = 6	x_2 = -3

Substituting the Values

To find the values of y, the values of x_1 and x_2 are substituted into either equation of the system. It is easier to substitute the values into Equation (II) because it is a linear equation, instead of quadratic.

y=5x -1
y_1 = 5( 6) - 1	y_2 = 5( -3) - 1
y_1 = 29	y_2 = -16

This indicates that the solutions of the system are the points (6,29) and (-3,-16).

Example

Skydive Recording

Finally, all the math and practice is making sense. Emily is now ready for her first jump! Instructor Sky Flyer prepared an equation to record Emily's height above the ground.

Her height from the ground when she jumps from the plane, measured in feet, is given by the following quadratic equation. h(t) = - 16 t^2 + 100t + 10 000 Again, h is Emily's height from the floor, while t is the time in seconds since Emily jumped. Using a linear equation, the instructor calculated Emily's height above the ground after her parachute was released. h(t) = - 16 t + 6700

a After how many seconds does Emily release her parachute? Use substitution to find the answer. Round the time to two decimal places.

b At what height does Emily release her parachute? Round the height to two decimal places.

Hint

a Which variable is most convenient to isolate?

b Substitute the answer from Part A into either equation.

Solution

a When Emily releases her parachute, both heights have to be equal. Therefore, to find the time is the same as finding the solution of the following nonlinear system.

h = - 16 t^2 +100t + 10 000 & (I) h = - 16 t + 6700 & (II) The first step to solve the system using substitution is to find a variable to isolate in one equation. In this case, the variable h is already isolated in both equations, so it is a convenient choice. h = - 16 t^2 +100t + 10 000 & (I) h = - 16 t + 6700 & (II) The value of h from Equation (II) can be substituted into Equation (I).

h = - 16 t^2 +100t + 10 000 & (I) h = - 16 t + 6700 & (II)

(I): h= - 16 t + 6700

- 16 t + 6700 = - 16 t^2 +100t + 10 000 h = - 16 t + 6700

(I): LHS-- 16 t + 6700=RHS-- 16 t + 6700

0 = - 16 t^2 +100t + 10 000 - (- 16 t + 6700) h = - 16 t + 6700

(I): Subtract terms

0 = -16t^2 + 116t + 3300 h = - 16 t + 6700

The quadratic equation obtained only depends on t. This quadratic equation is already written in standard form. at^2 + b t + c = 0 ⇓ -16 t^2 + 116 t + 3300 = 0 This equation can be solved using the quadratic formula.

t=- b±sqrt(b^2 - 4ac)/2a

Substitute values

t=-116±sqrt(116^2 - 4( -16)( 3300))/2( -16)

Calculate power

t=- 116±sqrt(13 456 - 4(-16)(3300))/2(-16)

t=- 116±sqrt(13 456 + 211 200)/-32

Add terms

t=- 116±sqrt(224 656)/- 32

There are two possible values for the time t.

t=- 116±sqrt(224 656)/- 32
t = -116 + sqrt(224 656)/-32	t = -116 - sqrt(224 656)/-32
t ≈ -11.17	t ≈ 18.44

The time after Emily jumps has to be positive. Because of this, the only solution that makes sense is the positive one. Therefore, Emily released her parachute about 18.44 seconds after she jumped.

b To find the height at which Emily released her parachute, the value from Part A can be substituted into either equation. In this case, the value will be substituted into Equation (II) because it is linear. Since the value of t is rounded, the equality is substituted with an approximation symbol.

h = -16t + 6700

t= 18.44

h ≈ -16* 18.44 + 6700

Add terms

h ≈ 6404.96

Therefore, the height at which Emily released her parachute is about 6404.96 feet.

Example

Planning New Formations

As Emily is on cloud nine after skydiving, she dreams of making a formation with other skydivers, where the skydivers make two intersecting circles.

She then realizes that Instructor Sky Flyer would tell her to solve a nonlinear system.

The formation she wants to make consists of two groups that each make a circle then intersect mid-air. This formation is modeled by the equations of two circles. (x+2)^2 + y^2 = 9 & (I) (x-4)^2 + y^2 = 9 & (II) Emily wants to find the points in which these circles intersect. This will help her have an idea of the roles they will have with the formation. Help Emily with her task and determine the correct pair of points of intersection. Solve using substitution.

Hint

Isolate a term of one equation to substitute into the other one.

Solution

To find the points of intersection is the same as solving the system. Neither equation of the system is linear, so this is a nonlinear system. The first step to solve this system using substitutions is to find a variable term that can be isolated. In this case, the convenient term to isolate is y^2. (x+2)^2 + y^2 = 9 & (I) (x-4)^2 + y^2 = 9 & (II) This term is isolated in Equation (II) to be substituted into Equation (I).

(x+2)^2 + y^2 = 9 & (I) (x-4)^2 + y^2 = 9 & (II)

(II): LHS-(x-4)^2=RHS-(x-4)^2

(x+2)^2 + y^2 = 9 & (I) y^2 = 9 - (x-4)^2 & (II)

(I): y^2= 9 - (x-4)^2

(x+2)^2 + 9 - (x-4)^2 = 9 & (I) y^2 = 9 - (x-4)^2 & (II)

The quadratic equation obtained only depends on x. To solve this equation, first the parenthesis need to be expanded.

(x+2)^2 + 9 - (x-4)^2 = 9

▼

Solve for x

LHS-9=RHS-9

(x+2)^2 - (x-4)^2 = 0

ExpandPosPerfectSquare

(a+b)^2=a^2+2ab+b^2

x^2 + 4x + 4 - (x-4)^2 = 0

(a-b)^2=a^2-2ab+b^2

x^2 + 4x + 4 - (x^2-8x + 16) = 0

Subtract terms

12x - 12 = 0

AddEqn

LHS+12=RHS+12

12x = 12

DivEqn

.LHS /12.=.RHS /12.

x = 1

This indicates that the circles intersect at x=1. To find the y-coordinate of the intersection points, the value x=1 has to be substituted into either equation. The equation when y^2 was isolated can be used.

y^2 = 9 - (x-4)^2

x= 1

y^2 = 9 - ( 1-4)^2

Subtract terms

y^2 = 9 - (-3 )^2

Calculate power

y^2 = 9 - 9

Subtract terms

y^2 = 0

Calculate root

y = 0

Since the only square root of 0 is 0, the only point of intersection is (1,0).

Closure

Finding the Intersections with a Circle

In this lesson's challenge, an equation of a circle centered at the origin with a radius of 2sqrt(2) was given. x^2 + y^2 = 8 The challenge was to find the points of intersection of that circle.

a Given the line y = 12x - 1, at what points does it intersect with the circle? Round the answer to one decimal place if needed.

b Given the parabola y = 12x^2, determine at what points it intersects with the circle. Round the answer to one decimal place if needed.

Hint

a Consider using substitution.

b Isolate the x^2 terms on both equations.

Solution

a The method used to find the points of intersection of the circle and the line is the same as finding the solution of the nonlinear system made by combining these equations.

x^2 + y^2 = 8 & (I) y = 1/2x - 1 & (II) The y variable is already isolated in Equation (II), so this can be substituted into the y term in Equation (I).

x^2 + y^2 = 8 & (I) y = 1/2x - 1 & (II)

(I): y= 1/2x - 1

x^2 + ( 1/2x - 1)^2 = 8 y = 1/2x - 1

(I): (a-b)^2=a^2-2ab+b^2

x^2 + 1/4x^2 - x + 1 = 8 y = 1/2x - 1

(I): Add terms

5/4x^2 - x + 1 = 8 y = 1/2x - 1

(I): Subtract terms

5/4x^2 - x -7 = 0 y = 1/2x - 1

It can be seen that the obtained quadratic equation is written in standard form. ax^2 + bx + c = 0 [0.5em] ⇓ 5/4x^2 + ( -1)x + ( -7) = 0 This equation can then be solved using the quadratic formula.

x=- b±sqrt(b^2 - 4ac)/2a

Substitute values

x=- ( -1)±sqrt(( -1)^2 - 4( 54)( -7))/2( 54)

▼

Simplify right-hand side

Calculate power

x=-(-1)±sqrt(1 - 4( 54)(-7))/2( 54)

x=-(-1)±sqrt(1 +35)/52

- (- a)=a

x=1±sqrt(1 +35)/52

Add terms

x=1±sqrt(36)/52

Calculate root

x = 1± 6/52

ExpandFrac

a/b=a * 2/b * 2

x= 2± 12/5

There are two possible values for x. These values change if adding or subtracting.

x = 2± 12/5
x = 2+ 12/5	x = 2- 12/5
2.8	-2

Finally, these values are substituted into Equation (II) to find the values of y of the points of intersection.

y = 12x - 1
y = 12( 2.8) - 1	y = 12( -2) - 1
0.4	-2

The points of intersection are (2.8,0.4) and (-2,-2).

b Finding the points of intersection between the circle and a parabola is achieved by finding the solutions of the nonlinear system.

x^2 + y^2 = 8 & (I) y = 1/2x^2 & (II) The x^2 term can be isolated on both equations.

x^2 + y^2 = 8 & (I) y = 1/2x^2 & (II)

(I): LHS-y^2=RHS-y^2

x^2 = 8 - y^2 y = 1/2x^2

(II): Rearrange equation

x^2 = 8 - y^2 1/2x^2 = y

MultEqn

(II): LHS * 2=RHS* 2

x^2 = 8 - y^2 x^2 = 2y

The Equation (II) can be subtracted from Equation (I) to solve the system using elimination.

x^2 = 8 - y^2 x^2 = 2y

(I): Subtract (II)

x^2 - x^2 = 8 - y^2 - 2y x^2 = 2y

(I): Subtract terms

0 = 8 - y^2 - 2y x^2 = 2y

(I): Rearrange equation

- y^2 - 2y + 8 = 0 x^2 = 2y

A quadratic equation of y is obtained. It can be noted that the equation is written in standard form. ax^2 + bx + c = 0 ⇓ ( -1) y^2 + ( -2)y + 8 = 0 This equation can be solved using the quadratic formula.

y=- b±sqrt(b^2 - 4ac)/2a

Substitute values

y=- ( -2)±sqrt(( -2)^2 - 4( -1)( 8))/2( -1)

Calculate power

y=-(-2)±sqrt(4 - 4(-1)(8))/2(-1)

y=-(-2)±sqrt(4 +32)/-2

- (- a)=a

y=2±sqrt(4 +32)/-2

Add terms

y=2±sqrt(36)/-2

Calculate root

y = 2 ± 6/-2

Now the two possible values for y can be found by adding or subtracting.

y = 2 ± 6/-2
y = 2 + 6/-2	y = 2 - 6/-2
-4	2

Now two values of y were obtained, and it should be noted that the value of y in Equation (II) has to be greater than or equal to 0. Therefore, the valid solution is 2, while -4 is an extraneous solution. The value of 2 can be substituted into either equation to find the possible values of x. In this case, it will be substituted into Equation (II).

y = 1/2x^2

y= 2

2 = 1/2x^2

MultEqn

LHS * 2=RHS* 2

4 = x^2

Rearrange equation

x^2 = 4