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Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
Chapter Review
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Exercise 55 Page 606

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

(12, 123) and ( -8,3)

Practice makes perfect

To solve the given system of equations, we can use the Substitution Method. y=x^2+2x-45 & (I) y=6x+51 & (II) The y-variable is isolated in both equations. This allows us to substitute its value 6x+51 for y in Equation (I).

y=x^2+2x-45 y=6x+51
Substitute

(I): y= 6x+51

6x+51=x^2+2x-45 y=6x+51
â–¼
(I): Simplify
SubEqn

(I): LHS-6x=RHS-6x

51=x^2-4x-45 y=6x+51
SubEqn

(I): LHS-51=RHS-51

0=x^2-4x-96 y=6x+51
RearrangeEqn

(I): Rearrange equation

x^2-4x-96=0 y=6x+51

Notice that in Equation (I), we have a quadratic equation in terms of only the x-variable. x^2-4x-96=0 ⇔ 1x^2+( - 4)x+( - 96)=0

We can substitute a= 1, b= - 4, and c= - 96 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a
SubstituteValues

Substitute values

x=- ( - 4)±sqrt(( - 4)^2-4( 1)( -96))/2( 1)
â–¼
Solve for x
NegNeg

- (- a)=a

x=4±sqrt((- 4)^2-4(1)(- 96))/2(1)
CalcPow

Calculate power

x=4±sqrt(16-4(1)(- 96))/2(1)
MultByOne

a * 1=a

x=4±sqrt(16-4(- 96))/2
MultNegNegOnePar

- a(- b)=a* b

x=4±sqrt(16+384)/2
AddTerms

Add terms

x=4±sqrt(400)/2
CalcRoot

Calculate root

x=4± 20/2

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=4± 20/2
x_1=4+20/2 x_2=4-20/2
x_1=24/2 x_2=-16/2
x_1=12 x_2=- 8

Now, consider Equation (II). y=6x+51 We can substitute x=12 and x=- 8 into the above equation to find the values for y. Let's start with x=12.

y=6x+51
Substitute

x= 12

y=6( 12)+51
â–¼
Solve for y
Multiply

Multiply

y=72+51
AddTerms

Add terms

y=123

We found that y=123 when x=12. One solution of the system, which is a point of intersection of the two graphs, is (12,123). To find the other solution, we will substitute -8 for x in Equation (II) again.

y=6x+51
Substitute

x= -8

y=6( -8)+51
â–¼
Solve for y
MultPosNeg

a(- b)=- a * b

y=-48 + 51
AddTerms

Add terms

y=3

We found that y=3 when x=-8. Therefore, our second solution, which is the other point of intersection of the two graphs, is ( - 8,3).

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arrow_right Exercises p.603-606
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Exercises 10 (Page 604)
Exercises 11 (Page 604)
Exercises 12 (Page 604)
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Exercises 47 (Page 606)
Exercises 48 (Page 606)
Exercises 49 (Page 606)
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Exercises 51 (Page 606)
Exercises 52 (Page 606)
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Exercises 54 (Page 606)
Exercises 55 (Page 606)
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Exercises 58 (Page 606)
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