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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 20 Page 599

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

(-7,100) and (-10,130)

Practice makes perfect

We want to solve the given system of equations using the Substitution Method. y=x^2+7x+100 & (I) y+10x=30 & (II) The y-variable is isolated in Equation (I). This allows us to substitute its value x^2+7x+100 for y in Equation (II).

y=x^2+7x+100 & (I) y+10x=30 & (II)

(II): y= x^2+7x+100

y=x^2+7x+100 x^2+7x+100+10x=30

(II): Add terms

y=x^2+7x+100 x^2+17x+100=30

(II): LHS-30=RHS-30

y=x^2+7x+100 x^2+17x+70=0

Notice that in Equation (II), we have a quadratic equation in terms of only the x-variable. x^2+17x+70=0 ⇔ 1x^2+ 17x+ 70=0

We can substitute a= 1, b= 17, and c= 70 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 17±sqrt(17^2-4( 1)( 70))/2( 1)

▼

Simplify right-hand side

Calculate power

x=-17±sqrt(289-4(1)(70))/2(1)

a * 1=a

x=-17±sqrt(289-4(70))/2

(- a)b = - ab

x=-17±sqrt(289-280)/2

Subtract term

x=-17±sqrt(9)/2

Calculate root

x=-17± 3/2

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=-17± 3/2
x_1=-17+ 3/2	x_2=-17- 3/2
x_1=-14/2	x_2=-20/2
x_1=-7	x_2=-10

Now, consider Equation (I). y=x^2+7x+100 We can substitute x=-7 and x=-10 into the above equation to find the values for y. Let's start with x=-7.

y=x^2+7x+100

x= -7

y=( -7)^2+7( -7)+100

▼

Solve for y

Calculate power

y=49+7(-7)+100

a(- b)=- a * b

y=49-49+100

Add and subtract terms

y=100

We found that y=100 when x=-7. One solution of the system, which is a point of intersection of the parabola and the line, is (-7,100). To find the other solution, we will substitute -10 for x in Equation (I) again.

y=x^2+7x+100

x= -10

y=( -10)^2+7( -10)+100

▼

Solve for y

Calculate power

y=100+7(-10)+100

a(- b)=- a * b

y=100-70+100

Add and subtract terms

y=130

We found that y=130 when x=-10. Therefore, our second solution, which is the other point of intersection of the parabola and the line, is (-10,130).

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 (-7,100). We will substitute -7 and 100 for x and y, respectively, in Equation (I) and Equation (II).

y=x^2+7x+100 & (I) y+10x=30 & (II)

(I), (II): x= -7, y= 100

100? =( -7)^2+7( -7)+100 100+10( -7)? =30

▼

Simplify

(I): Calculate power

100? =49+7(-7)+100 100+10(-7)? =30

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

100? =49-49+100 100-70? =30

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

100=100 ✓ 30=30 ✓

Since both equations produced true statements, the solution (-7,100) is correct. Let's now check (-10,130).

y=x^2+7x+100 & (I) y+10x=30 & (II)

(I), (II): x= -10, y= 130

130? =( -10)^2+7( -10)+100 130+10( -10)? =30

▼

Simplify

(I): Calculate power

130? =100+7(-10)+100 130+10(-10)? =30

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

130? =100-70+100 130-100? =30

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

130=130 ✓ 30=30 ✓

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