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McGraw Hill Integrated II, 2012 View details

5. Solving Quadratic Equations by Using the Quadratic Formula

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Exercise 49 Page 138

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

MultPar

Multiply parentheses

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

SubTerm

Subtract term

500-190x+14x^2= 375

CommutativePropAdd

Commutative Property of Addition

14x^2-190x+500=375

SubEqn

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

NegNeg

- (- a)=a

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

CalcPow

Calculate power

x=190±sqrt(36 100-4(14)(125))/2(14)