SOLUTION: The length of a rectangle is 6 more inches than its width. The area of the rectangle is 91 square inches. Find the dimensions of the rectangle.
Suppose width is x inches than le
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Suppose width is x inches than le
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Question 756044: The length of a rectangle is 6 more inches than its width. The area of the rectangle is 91 square inches. Find the dimensions of the rectangle.
Suppose width is x inches than length is x+6
Area of a rectangle is= l*b
91=(x)*(x+6)
91= x^2+6x
0=x^2+6x-91
0=x^2+13x-7x-91
0=x(x+13)-7(x+13)
0=(x-7)(x+13)
0=(x-7) OR 0=(x+13)
7=x OR -13=x (negative width not possible)
x=7 inches width and length is six more so 7+6=13
Answer is 7 and 13 are the length and width of the rectangle
I hope u find my solution useful Answer by nihar*2013(34) (Show Source):
You can put this solution on YOUR website! The length of a rectangle is 6 more inches than its width. The area of the rectangle is 91 square inches. Find the dimensions of the rectangle.
Suppose width is x inches than length is x+6
Area of a rectangle is= l*b
91=(x)*(x+6)
91= x^2+6x
0=x^2+6x-91
0=x^2+13x-7x-91
0=x(x+13)-7(x+13)
0=(x-7)(x+13)
0=(x-7) OR 0=(x+13)
7=x OR -13=x (negative width not possible)
x=7 inches width and length is six more so 7+6=13
Answer is 7 and 13 are the length and width of the rectangle
I hope u find my solution useful
