Quadratic applications
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Quadratic Applications. Quadratic Applications. Exercise 1a ) Sandy’s bedroom is in the shape of a rectangle that has an area of 120 square feet. The width is two feet less than the length. Find the dimensions of Sandy’s room. 1) Legend:. Let x =. Length . x-2 =. width.

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Quadratic Applications

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Quadratic applications

Quadratic Applications


Quadratic applications

Quadratic Applications

Exercise 1a) Sandy’s bedroom is in the shape of a rectangle that has an area of 120 square feet. The width is two feet less than the length. Find the dimensions of Sandy’s room.

1) Legend:

Let x =

Length

x-2 =

width

Notice: We came up with two answers….how can this be?

2) Draw a picture:

We don’t really have two answers. We are talking about length, so all our answer must be positive. So we get to eliminate the negative answers.

A=120 sq ft

x

x-2

3) Formula:

Area = length * width

4) Equation:

So……..x=12

x(x-2)=120

Go back to the legend and calculate the width.

x2-2x-120=0

(x-12)(x+10)=0

The length is 12 feet and the width is 10 feet.

(x-12)=0 or (x+10)=0

x=12 or x = -10


Quadratic applications

Quadratic Applications

Exercise 2a) LaTonya’s lawn is in the shape of a right triangle. The area of the lawn is 20 square yards. If the shorter leg o f the triangle is 3 yards less than the longer leg, what are the lengths of the legs of the triangular lawn?

1) Legend:

Let x =

Long leg

x-3 =

Short leg

Eliminate the negative answer.

2) Draw a picture:

x

So……..x=8

A=20 sq yrd

Go back to the legend and calculate the short leg.

x-3

3) Formula:

The long leg is 8 yards and the short leg is 5 yards.

Area = ½ base* height

4) Equation:

x2 - 3x -40=0

1x(x-3)=20

2

(x-8) (x+5)=0

x2-3x=20

2

(x-8)=0 or (x+5)=0

x=8 or x = -5

x2-3x = 40


Quadratic applications

Quadratic Applications

Exercise 3a) Shay’s garden is in the shape of a right triangle. The hypotenuse is 15 feet, and the longer leg is 3 feet more than the shorter leg. What are the lengths of the two legs?

1) Legend:

Let x =

short leg

x+3 =

Long leg

Eliminate the negative answer.

2) Draw a picture:

15 ft

So……..x=9

x

Go back to the legend and calculate the long leg.

x+3

3) Formula:

or leg2 + leg2 = hypotenuse2

a2 + b2 = c2

4) Equation:

The short leg is 9 feet and the long leg is 12 feet.

x2+(x+3)2=152

2(x-9) (x+12)=0

x2+x2+6x+9=225

(x-9)=0 or (x+12)=0

2x2 - 6x - 216 = 0

x=9 or x = -12

2(x2 - 3x -108)=0


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