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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

8. Systems of Linear and Quadratic Equations

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Exercise 19 Page 599

To solve the equation ax^2+bx+c=0, use the Quadratic Formula.

(- 9, - 47) and (6, - 2)

Practice makes perfect

We want to solve the given system of equations using the Substitution Method. y=3x-20 & (I) y=- x^2+34 & (II) The y-variable is isolated in Equation (II). This allows us to substitute its value - x^2+34 for y in Equation (I).

y=3x-20 y=- x^2+34

(I): y= - x^2+34

- x^2+34=3x-20 y=- x^2+34

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(I): Simplify

(I): LHS+20=RHS+20

- x^2+54=3x y=- x^2+34

(I): LHS-3x=RHS-3x

- x^2-3x+54=0 y=- x^2+34

Notice that in Equation (I) we have a quadratic equation in terms of only the x-variable. - x^2-3x+54=0 ⇔ - 1x^2+( - 3)x+ 54=0

We can substitute a= - 1, b= - 3, and c= 54 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 3)±sqrt(( - 3)^2-4( - 1)( 54))/2( - 1)

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Solve for x

- (- a)=a

x=3±sqrt((- 3)^2-4(- 1)(54))/2(- 1)

Calculate power

x=3±sqrt(9-4(- 1)(54))/2(- 1)

Multiplication Property of -1

x=3±sqrt(9+4(54))/- 2

Multiply

x=3±sqrt(9+216)/- 2

Calculate root

x=3 ± 15/- 2

This result tells us that we have two solutions for x. One of them will use the positive sign and the other one will use the negative sign.

x=3 ± 15/- 2
x_1=3+15/- 2	x_2=3-15/- 2
x_1=18/- 2	x_2=- 12/- 2
x_1=- 9	x_2=6

Now, consider Equation (II). y=- x^2+34 We can substitute x=- 9 and x=6 into the above equation to find the values for y. Let's start with x=- 9.

y=- x^2+34

x= - 9

y=- ( - 9)^2+34

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Solve for y

Calculate power

y=-81+34

Subtract term

y=- 47

We found that y=- 47 when x=- 9. One solution of the system, which is a point of intersection of the two parabolas, is (- 9,- 47). To find the other solution we will substitute 6 for x in Equation (II) again.

y=- x^2+34

x= 6

y=- 6^2+34

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Solve for y

Calculate power

y=- 36+34

Add terms

y=- 2

We found that y=- 2 when x=6. Therefore, our second solution, which is the other point of intersection of the two parabolas, is ( 6,- 2).

Checking Our Answer

Checking the answer

We can check our answers by substituting the points into both equations. If they produce true statements, our solutions are correct. Let's start by checking (- 9, - 47). We will substitute - 9 and - 47 for x and y, respectively, in Equation (I) and Equation (II).

y=3x-20 y=- x^2+34

(I), (II): x= - 9, y= - 47

- 47? =3( - 9)-20 - 47? =- ( - 9)^2+34

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Simplify

(I): a(- b)=- a * b

- 47? =- 27-20 - 47? =- (- 9)^2+34

NegBaseToPosBase

(II): (- a)^2=a^2

- 47? =- 27-20 - 47? =- 81+34

(I), (II): Add and subtract terms

- 47=- 47 ✓ - 47=- 47 ✓

Since both equations produced true statements, the solution (- 9,- 47) is correct. Let's now check ( 6, - 2).

y=3x-20 y=- x^2+34

(I), (II): x= 6, y= - 2

- 2? =3( 6)-20 - 2? =- ( 6)^2+34

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Simplify

(I): Multiply

- 2? =18-20 - 2? =- (6)^2+34

(II): Calculate power

- 2? =18-20 - 2? =- 36+34

(I), (II): Add and subtract terms

- 2=- 2 ✓ - 2=- 2 ✓

Since again both equations produce true statements, the solution (6, - 2) is also correct.

Subchapter links

Lesson Check p.599

Practice and Problem-Solving Exercises p.599-600

Standardized Test Prep p.601

Mixed Review p.601