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McGraw Hill Glencoe Algebra 1, 2012

MH

McGraw Hill Glencoe Algebra 1, 2012 View details

5. Solving Quadratic Equations by Using the Quadratic Formula

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Exercise 49 Page 588

Practice makes perfect

a We want to write an equation for the area of the section with text. The dimensions of this section are ( 20-2x) inches and ( 25-7x)inches as shown below.

Bartolo wants the area of this section to be three-fourths of the original area, so it has an area of 375in.^2. 3/4 * ( 20* 25) = 375 One side of the equation is 375, the other side is the expression for the area of the section, the width times the length. ( 20-2x)( 25-7x) = 375

b We need to rewrite the equation in standard form to be able to use the Quadratic Formula.

(20-2x)(25-7x) = 375

▼

Multiply parentheses

Multiply parentheses

500-140x-50x+14x^2= 375

Subtract term

500-190x+14x^2= 375

CommutativePropAdd

Commutative Property of Addition

14x^2-190x+500=375

LHS-375=RHS-375

14x^2-190x+125=0

Now, we can identify the values of a, b, and c.

14x^2-190x+125=0 ⇕ 14x^2+( - 190)x+ 125=0 We see that a= 14, b= - 190, and c= 125. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 190)±sqrt(( - 190)^2-4( 14)( 125))/2( 14)

▼

Simplify right-hand side

- (- a)=a

x=190±sqrt((- 190)^2-4(14)(125))/2(14)

Calculate power

x=190±sqrt(36 100-4(14)(125))/2(14)

Multiply

x=190±sqrt(36 100-4(14)(125))/28

Multiply

x=190±sqrt(36 100-4(1750))/28

(- a)b = - ab

x=190±sqrt(36 100-7000)/28

Subtract term

x=190±sqrt(29 100)/28

The solutions for this equation are x= 190 ± sqrt(29 100)28. Let's separate them into the positive and negative cases.

x=190 ± sqrt(29 100)/28
x_1=190 + sqrt(29 100)/28	x_2=190 - sqrt(29 100)/28
x_1=190/28 + sqrt(29 100)/28	x_2=190/28 - sqrt(29 100)/28
x_1 ≈ 12.9	x_2 ≈ 0.7

Using the Quadratic Formula, we found that the solutions of the given equation are x_1≈ 12.9 and x_2≈ 0.7.

c The margins of the poster should be 0.7 in. on the sides, 2.8 in. on the top, and 2.1 in. on the bottom.

x & = 0.7 4x &= 4(0.7)= 2.8 3x &=3(0.7)= 2.1 Note that the other solution x_1 ≈ 12.9, makes the lengths negative.

Subchapter links

Check Your Understanding p.587

Practice and Problem Solving p.587-588

H.O.T. Problems p.588

Standardized Test Practice p.589

Spiral Review p.589

Skills Review p.589