SOLUTION: This was my innitial problem;
"The are of a rectangle is 165 ft^2. The Length is 4 ft greater than the width. Find the dimensions of the rectangle."
This my question;
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-> SOLUTION: This was my innitial problem;
"The are of a rectangle is 165 ft^2. The Length is 4 ft greater than the width. Find the dimensions of the rectangle."
This my question;
After
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Question 540495: This was my innitial problem;
"The are of a rectangle is 165 ft^2. The Length is 4 ft greater than the width. Find the dimensions of the rectangle."
This my question;
After you simplify the word problem into a Quadratic Equation (the process below)..
W*L=165, L=W+4
W(W+4)=165
W^2+4W-165=0
How do you find this next part;
(W-11)(W+15)=0
W-11=0
How did the tutor figure 11 and 15? Is it because of the factoring? Perfect squares? Please help.
You can put this solution on YOUR website! width = x
length = x+4
x(x+4)=165
x^2+4x-165=0
Split the middle term such that when you add the split terms you get -4
and when you multiply you get -165
Take the prime factors of 165
3*5*11
3*5 = 15
you need 4
+15-11=-4
x^2+15x-11x-165=0
x(x+15)-11(x+15)=0
(x+15)(x-11)=0
take the positive value since width cannot be negative
x=11
length= 11+4=15
Thats how your teacher got 11 & 15
m.ananth@hotmail.ca
