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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 22 Page 599

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

(2,8) and (-1/2,1/2)

Practice makes perfect

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

3x-y=-2 & (I) 2x^2=y & (II)

(I): y= 2x^2

3x- 2x^2=-2 2x^2=y

(I): LHS+2x^2=RHS+2x^2

3x=2x^2-2 2x^2=y

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

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

(I): Rearrange equation

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

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

We can substitute a= 2, b= -3, and c= -2 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

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

▼

Simplify right-hand side

- (- a)=a

x=3±sqrt((-3)^2-4(2)(-2))/2(2)

Calculate power

x=3±sqrt(9-4(2)(-2))/2(2)

(- a)b = - ab

x=3±sqrt(9-8(-2))/2(2)

MultNegNegOnePar

- a(- b)=a* b

x=3±sqrt(9+16)/2(2)

Add terms

x=3±sqrt(25)/2(2)

Multiply

x=3±sqrt(25)/4

Calculate root

x=3± 5/4

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=3± 5/4
x_1=3+ 5/4	x_2=3- 5/4
x_1=8/4	x_2=-2/4
x_1=2	x_2=-1/2

Now, consider Equation (II). 2x^2=y We can substitute x=2 and x=- 12 into the above equation to find the values for y. Let's start with x=2.

2x^2=y

x= 2

2( 2)^2=y

▼

Solve for y

Calculate power

2(4)=y

Multiply

8=y

Rearrange equation

y=8

We found that y=8 when x=2. One solution of the system, which is a point of intersection of the parabola and the line, is (2,8). To find the other solution, we will substitute - 12 for x in Equation (II) again.

2x^2=y

x= -1/2

2( -1/2)^2=y

▼

Solve for y

NegBaseToPosPow

(- a)^2 = a^2

2(1/2)^2=y

(a/b)^m=a^m/b^m

2(1/4)=y

MoveLeftFacToNumOne

a* 1/b= a/b

2/4=y

a/b=.a /2./.b /2.

1/2=y

Rearrange equation

y=1/2

We found that y= 12 when x=- 12. Therefore, our second solution, which is the other point of intersection of the parabola and the line, is (- 12, 12).

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 (2,8). We will substitute 2 and 8 for x and y, respectively, in Equation (I) and Equation (II).

3x-y=-2 & (I) 2x^2=y & (II)

(I), (II): x= 2, y= 8

3( 2)- 8? =-2 2( 2)^2? = 8

▼

Simplify

(II): Calculate power

3(2)-8? =-2 2(4)? =8

(I), (II): Multiply

6-8? =-2 8=8 ✓

(I): Subtract term

-2=-2 ✓ 8=8 ✓

Since both equations produce true statements, the solution (2,8) is correct. Let's now check (- 12, 12).

3x-y=-2 & (I) 2x^2=y & (II)

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

3( - 12)- 12? =-2 2( - 12)^2? = 12

▼

Simplify

NegBaseToPosPow

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

3(- 12)- 12? =-2 2( 12)^2? = 12

(II): (a/b)^m=a^m/b^m

3(- 12)- 12? =-2 2( 14)? = 12

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

- 3( 12)- 12? =-2 2( 14)? = 12

(I), (II): a* 1/b= a/b

- 32- 12? =-2 24? = 12

(I): Subtract fractions

- 42? =-2 24? = 12

(I): Calculate quotient

- 2=-2 ✓ 24? = 12

(II): a/b=.a /2./.b /2.

-2=-2 ✓ 12= 12 ✓

Since again both equations produced true statements, the solution (-10,130) 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