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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 18 Page 599

Use the Quadratic Formula to solve the equation of the form ax^2+bx+c=0.

(8,42) and (-2,2)

Practice makes perfect

We want to solve the given system of equations using the Substitution Method. y=x^2-2x-6 & (I) y=4x+10 & (II) The y-variable is isolated in Equation (I). This allows us to substitute its value x^2-2x-6 for y in Equation (II).

y=x^2-2x-6 & (I) y=4x+10 & (II)

(II): y= x^2-2x-6

y=x^2-2x-6 x^2-2x-6=4x+10

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

y=x^2-2x-6 x^2-6x-6=10

(II): LHS-10=RHS-10

y=x^2-2x-6 x^2-6x-16=0

Notice that in Equation (II), we have a quadratic equation in terms of only the x-variable. x^2-6x-16=0 ⇔ 1x^2+( - 6)x+( -16)=0

We can substitute a= 1, b= - 6, and c= -16 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 6)±sqrt(( - 6)^2-4( 1)( -16))/2( 1)

▼

Simplify right-hand side

- (- a)=a

x=6±sqrt((- 6)^2-4(1)(-16))/2(1)

Calculate power

x=6±sqrt(36-4(1)(-16))/2(1)

a * 1=a

x=6±sqrt(36-4(-16))/2

MultNegNegOnePar

- a(- b)=a* b

x=6±sqrt(36+64)/2

Add terms

x=6±sqrt(100)/2

Calculate root

x=6± 10/2

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

x=6± 10/2
x_1=6+ 10/2	x_2=6- 10/2
x_1=16/2	x_2=-4/2
x_1=8	x_2=-2

Now, consider Equation (I). y=x^2-2x-6 We can substitute x=8 and x=-2 into the above equation to find the values for y. Let's start with x=8.

y=x^2-2x-6

x= 8

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

▼

Solve for y

Calculate power

y=64-2(8)-6

Multiply

y=64-16-6

Subtract terms

y=42

We found that y=42 when x=8. One solution of the system, which is a point of intersection of the parabola and the line, is (8,42). To find the other solution, we will substitute -2 for x in Equation (I) again.

y=x^2-2x-6

x= -2

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

▼

Solve for y

Calculate power

y=4-2(-2)-6

MultNegNegOnePar

- a(- b)=a* b

y=4+4-6

Add and subtract terms

y=2

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

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 (8,42). We will substitute 8 and 42 for x and y, respectively, in Equation (I) and Equation (II).

y=x^2-2x-6 & (I) y=4x+10 & (II)

(I), (II): x= 8, y= 42

42? = 8^2-2( 8)-6 42? =4( 8)+10

▼

Simplify

(I): Calculate power

42? =64-2(8)-6 42? =4(8)+10

(I), (II): Multiply

42? =64-16-6 42? =32+10

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

42=42 ✓ 42=42 ✓

Since both equations produced true statements, the solution (8,42) is correct. Let's now check (-2,2).

y=x^2-2x-6 & (I) y=4x+10 & (II)

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

2? =( -2)^2-2( -2)-6 2? =4( -2)+10

▼

Simplify

(I): Calculate power

2? =4-2(-2)-6 2? =4(-2)+10

MultNegNegOnePar

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

2? =4+4-6 2? =4(-2)+10

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

2? =4+4-6 2? =-8+10

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

2=2 ✓ 2=2 ✓

Since again both equations produce true statements, the solution (-2,2) 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