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Pearson Algebra 1 Common Core, 2011 View details

Chapter Review

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Exercise 58 Page 606

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

(-13, 64) and ( 3,-16)

Practice makes perfect

To solve the given system of equations we can use the Substitution Method. y=x^2+5x-40 & (I) y+1=-5 x & (II) ⇓ y=x^2+5x-40 & (I) y=-5 x-1 & (II) The y-variable is already isolated in Equation (I). Note that, we can also isolate it in Equation (II) by subtracting 1 from both sides of the equation. This allows us to substitute its value -5 x-1 for y in Equation (I).

y=x^2+5x-40 y=-5 x-1

Substitute

(I): y= -5 x-1

-5 x-1=x^2+5x-40 y=-5 x-1

▼

(I): Simplify

AddEqn

(I): LHS+1=RHS+1

-5 x=x^2+5x-39 y=-5 x-1

AddEqn

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

0=x^2+10x-39 y=-5 x-1

RearrangeEqn

(I): Rearrange equation

x^2+10x-39=0 y=-5 x-1

Notice that in Equation (I), we have a quadratic equation in terms of only the x-variable. x^2+10x-39=0 ⇔ 1x^2+ 10x+( -39)=0 We can substitute a= 1, b= 10, and c= -39 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( 10)±sqrt(( 10)^2-4( 1)( -39))/2( 1)

▼

Solve for x

CalcPow

Calculate power

x=-10±sqrt(100-4(1)(-39))/2(1)

MultByOne

a * 1=a

x=-10±sqrt(100-4(-39))/2

MultNegNegOnePar

- a(- b)=a* b

x=-10±sqrt(100+156)/2

AddTerms

Add terms

x=-10±sqrt(256)/2

CalcRoot

Calculate root

x=-10± 16/2

FactorOut

Factor out 2

x=2(-5± 8)/2

CancelCommonFac

Cancel out common factors

x=-5 ±8

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=-5 ±8
x_1=-5 +8	x_2=-5 -8
x_1=3	x_2=-13

Now, consider Equation (II). y=-5 x-1 We can substitute x=3 and x=-13 into the above equation to find the values for y. Let's start with x=3.

y=-5 x-1

Substitute

x= 3

y=-5( 3)-1

▼

Solve for y

Multiply

y=-15-1

SubTerm

Subtract term

y=-16

We found that y=-16 when x=3. One solution of the system, which is a point of intersection of the two graphs, is (3,-16). To find the other solution, we will substitute -13 for x in Equation (II) again.

y=-5 x-1

Substitute

x= -13

y=-5( -13)-1

▼

Solve for y

MultNegNegOnePar

- a(- b)=a* b

y=65-1

SubTerm

Subtract term

y=64

We found that y=64 when x=-13. Therefore, our second solution, which is the other point of intersection of the two graphs, is ( -13,64).

Exercise 58 Page 606

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