Exercise 45 Page 548

We are given that the dimensions of a rectangle are y and y^2+1, and its perimeter is 14 units. We can write an equation using the fact that the perimeter of a rectangle is a twice the sum of its dimensions. 2[ y+( y^2+1)]= 14 Let's simplify the above equation. Since we ended with a quadratic equation, we will use the Quadratic Formula to solve it. ay^2+ by+ c=0 ⇔ y=- b± sqrt(b^2-4 a c)/2 a We first need to identify the values of a, b, and c. y^2+y-6=0 ⇔ 1y^2+ 1y+( -6)=0 We see that a= 1, b= 1, and c= -6. Let's substitute these values into the Quadratic Formula.

y=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

y=- 1±sqrt(1^2-4( 1)( -6))/2( 1)

▼

Solve for y and Simplify

BaseOne

1^a=1

y=-1±sqrt(1-4(1)(-6))/2(1)

MultByOne

a * 1=a

y=-1±sqrt(1-4(-6))/2

MultNegNegOnePar

- a(- b)=a* b

y=-1±sqrt(1+24)/2

AddTerms

Add terms

y=-1±sqrt(25)/2

CalcRoot

Calculate root

y=-1± 5/2

We will consider only positive case as y is a dimension of a rectangle.

y=-1+5/2

AddTerms

Add terms

y=4/2

CalcQuot

Calculate quotient

y=2

We got that one side of this rectangle is 2. By substituting y=2 into y^2+1, we will evaluate the second side.

y^2+1

Substitute

y= 2

2^2+1

CalcPow

Calculate power

4+1

AddTerms

Add terms

5

The longer side of this rectangle is 5. Now, we can evaluate the ratio of the longer side of the rectangle to the shorter side. The longer side/The shorter side=5/2 The ratio of the longer side to the shorter side is 5:2.