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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.

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Quadratic Applications

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  1. Quadratic Applications

  2. 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

  3. 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

  4. 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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