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Core Connections Algebra 2, 2013

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Core Connections Algebra 2, 2013 View details

2. Section 4.2

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Exercise 77 Page 190

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

(1,12) and (-5,42)

Practice makes perfect

We want to solve the given system of equations using the Substitution Method. y=x^2-x+12 & (I) y=2x^2+3x+7 & (II) The y-variable is isolated in both equations. This means we can choose the equation we will use for our initial substitution. Let's substitute x^2-x+12 for y in Equation (II).

y=x^2-x+12 y=2x^2+3x+7

(II): y= x^2-x+12

y=x^2-x+12 x^2-x+12=2x^2+3x+7

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(II): Simplify

(II): LHS-x^2=RHS-x^2

y=x^2-x+12 - x+12=x^2+3x+7

(II): LHS+x=RHS+x

y=x^2-x+12 12=x^2+4x+7

(II): LHS-12=RHS-12

y=x^2-x+12 0=x^2+4x-5

(II): Rearrange equation

y=x^2-x+12 x^2+4x-5=0

Notice that in Equation (II) we have a quadratic equation in terms of only the x-variable. x^2+4x-5=0 ⇔ 1x^2+ 4x+( -5)=0

Now, recall the Quadratic Formula. x=- b±sqrt(b^2-4ac)/2a We can substitute a= 1, b= 4, and c= -5 into this formula to solve the quadratic equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 4±sqrt(4^2-4( 1)( -5))/2( 1)

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Solve for x

Calculate power

x=-4±sqrt(16-4(1)(-5))/2(1)

a * 1=a

x=-4±sqrt(16-4(-5))/2

MultNegNegOnePar

- a(- b)=a* b

x=-4±sqrt(16+20)/2

Add terms

x=-4±sqrt(36)/2

Calculate root

x=-4±6/2

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±6/2
x_1=-4+6/2	x_2=-4-6/2
x_1=2/2	x_2=-10/2
x_1=1	x_2=-5

Now, consider Equation (I). y=x^2-x+12 We can substitute x=1 and x=-5 into the above equation to find the values for y. Let's start with x=1.

y=x^2-x+12

x= 1

y= 1^2- 1+12

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Solve for y

1^a=1

y=1-1+12

Add and subtract terms

y=12

We found that y=12 when x=1. One solution of the system, which is a point of intersection of the two parabolas, is (1,12). To find the other solution, we will substitute -5 for x in Equation (I) again.

y=x^2-x+12

x= -5

y=( -5)^2-( -5)+12

▼

Solve for y

Calculate power

y=25-(-5)+12

- (- a)=a

y=25+5+12

Add terms

y=42

We found that y=42 when x=-5. Therefore, our second solution, which is the other point of intersection of the two parabolas, is (-5,42).

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 (-5,42). We will substitute -5 and 42 for x and y, respectively, in Equation (I) and Equation (II).

y=x^2-x+12 y=2x^2+3x+7

(I), (II): x= -5, y= 42

42? =( -5)^2-( -5)+12 42? =2( -5)^2+3( -5)+7

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Simplify

(I), (II): Calculate power

42? =25-(-5)+12 42? =2(25)+3(-5)+7

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

42? =25+5+12 42? =2(25)+3(-5)+7

(II): Multiply

42? =25+5+12 42? =50-15+7

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

42=42 ✓ 42=42 ✓

Since both equations produce true statements, the solution (-5,42) is correct. Let's now check (1,12).

y=x^2-x+12 y=2x^2+3x+7

(I), (II): x= 1, y= 12

12? =( 1)^2- 1+12 12? =2( 1)^2+3( 1)+7

▼

Simplify

(I), (II): Calculate power

12? =1-1+12 12? =2(1)+3(1)+7

(II): a * 1=a

12? =1-1+12 12? =2+3+7

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

12=12 ✓ 12=12 ✓

Since again both equations produce true statements, the solution (1,12) is also correct.

Subchapter links

4.2.1 Solving Inequalities with One or Two Variables p.188-190

4.2.2 Using Systems to Solve a Problem p.193-194

4.2.3 Application of Systems of Linear Inequalities p.197-198

4.2.4 Using Graphs to Find Solutions p.200-201