QUADRATIC FUNCTIONS Word Problems
QUADRATIC FUNCTIONS (WORD PROBLEMS)
1. The area of a rectangle is 560 square inches. The length is 3 more than twice the width. Find the length and the width.
Representation:
Let L be the length and let W be the width.
The length is 3 more than twice the width, so
The area is 560, so
Equation:
Plug in and solve for W:
Solution:
Use the Quadratic Formula:
Since the width can't be negative, I get .
The length is
2. The hypotenuse of a right triangle is 4 times the smallest side. The third side is . Find the hypotenuse and the smallest side.
Representation:
Let s be the smallest side and let h be the hypotenuse. By Pythagoras,
The hypotenuse is 4 times the smallest side, so
Equation:
Plug into and
Solution:
Since doesn't make sense, the solution is .
Then
Remember:
Draw a picture and Write a “Let” statement
Write an equation
Solve the equation ( REMEMBER: YOU CAN’T HAVE A NEGATIVE LENGTH)
Check to see if your solution makes sense
Re-Read the problem to make sure you answered the question.
Solve for these problems. The answer is already provided as your guide.
1. The altitude of a triangle is 5 less than its base. The area of the triangle is 42 square inches. Find its base and altitude.
Answer: The base is 12 and the altitude is 7.
2. The length of a rectangle exceeds its width by 4 inches. Find the dimensions of the rectangle it its area is 96 square inches.
Answer: The width is 8 and the length is 12.
3. If the measure of one side of a square is increased by 2 centimeters and the measure of the adjacent side is decreased by 2 centimeters, the area of the resulting rectangle is 32 square centimeters. Find the measure of one side of the square.
Answer: The measure of one side of the square is 6.
Please visit this link for your review.
http://www.virtualnerd.com/algebra-1/quadratic-equations-functions/
1. The area of a rectangle is 560 square inches. The length is 3 more than twice the width. Find the length and the width.
Representation:
Let L be the length and let W be the width.
The length is 3 more than twice the width, so
The area is 560, so
Equation:
Plug in and solve for W:
Solution:
Use the Quadratic Formula:
Since the width can't be negative, I get .
The length is
2. The hypotenuse of a right triangle is 4 times the smallest side. The third side is . Find the hypotenuse and the smallest side.
Representation:
Let s be the smallest side and let h be the hypotenuse. By Pythagoras,
The hypotenuse is 4 times the smallest side, so
Equation:
Plug into and
Solution:
Since doesn't make sense, the solution is .
Then
Remember:
Draw a picture and Write a “Let” statement
Write an equation
Solve the equation ( REMEMBER: YOU CAN’T HAVE A NEGATIVE LENGTH)
Check to see if your solution makes sense
Re-Read the problem to make sure you answered the question.
Solve for these problems. The answer is already provided as your guide.
1. The altitude of a triangle is 5 less than its base. The area of the triangle is 42 square inches. Find its base and altitude.
Answer: The base is 12 and the altitude is 7.
2. The length of a rectangle exceeds its width by 4 inches. Find the dimensions of the rectangle it its area is 96 square inches.
Answer: The width is 8 and the length is 12.
3. If the measure of one side of a square is increased by 2 centimeters and the measure of the adjacent side is decreased by 2 centimeters, the area of the resulting rectangle is 32 square centimeters. Find the measure of one side of the square.
Answer: The measure of one side of the square is 6.
Please visit this link for your review.
http://www.virtualnerd.com/algebra-1/quadratic-equations-functions/