SOLUTION: I have having great difficulty with these worded problems; your help would be greatly appreciated. Thank you. It asks me: To solve the following word problem using quadratic equ

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Question 130984: I have having great difficulty with these worded problems; your help would be greatly appreciated. Thank you.
It asks me:
To solve the following word problem using quadratic equation:
Then goes on to state:

Given the area of a rectangle is 90m2 and the perimeter of the rectangle is 40m, find the breadth of the rectangle.

Thank you.
Anthony

Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website!
XY=90 OR Y=90/X
2X+2Y=40 NOW REPLACE Y WITH(90/X)
2X+2(90/X)=40
2X+180/X=40
(2X^2+180)/X=40
2X^2+180=40X
2X^2-40X+180=0
2(X^2-20X+90)=0
USING THE QUADRATIC EQUATION WE GET:

X=(20+-SQRT[-20^2-4*1*90])/2*1
X=(20+-SQRT[400-360])/2
X=(20+-SQRT40)/2
X=(20+-6.32)/2
X=(20+6.32)/2
X=26.32/2
X=13.16 ANSWER.
X=(20-6.32)/2
X=13.68/2
X=6.84 ANSWER.