Question 1187175: I'm not really understanding how to set up this factoring equation. The worksheet said to set up each word problem with an equation then solve by factoring and using the zero-product property with no guess and check. The word problem:
A diagonal of a rectangle is 5 cm longer than 4 times the width of the rectangle. Find the length of the rectangle if it is 1 cm less than the length of the diagonal.
Thank you!
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! A diagonal of a rectangle is 5 cm longer than 4 times the width of the rectangle. Find the length of the rectangle if it is 1 cm less than the length of the diagonal.
let W equal the width.
then 4W + 5 equals the diagonal.
then 4W + 4 equals the length.
the length and the width and the diagonal form a right triangle.
4W + 5 is the hypotenuse.
4W + 4 is one leg.
W is the other leg.
since the sum of the square of the legs is equal to the square of the hypotenuse, you get:
W^2 + (4W+4)^2 = (4W+5)^2
simplify that equation to get:
17W^2 + 32W + 16 = 16W^ + 40W + 25
subtract the right side of the equation from both sides of the equation to get:
W^2 - 8W - 9 = 0
that's your quadratic equation.
factor that to get:
(W - 9) * (W + 1) = 0
solve for W to get:
W = 9 or W = -1
confirm by replacing W with 9 in the original equation to get:
4W + 5 = 4 * 9 + 5 = 36 + 5 = 41 is the hypotenuse.
4W + 4 = 4 * 9 + 4 = 36 + 4 = 40 is the length of the rectangle.
9 is the width of the rectangle.
by pythagorus, length^2 + width^2 = diagonal^2.
you get 40^2 + 9^2 = 41^2.
this becomes 1681 = 1681, confirming the values of length and width are good.
your solution is that the length of the rectangle is equal to 40.
this is one less than the diagonal.
if there's any part of this you don't understand, then let me know and i'll take you through it.
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