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what are the dimensions of the rectangle if its area is equals to 119 sq/cm. while it's length is 3 more than it's width?


Sagot :

Area of a rectangle is A = LW
but L on that case is L=W+3 and A=119
A = LW
119 = (W +3)(W)
119 = W² + 3W
w² + 3W - 119 = 0
using quadratic formula
[tex]x = \frac{-b (+-) \sqrt{b^2 -4ac} }{2a} [/tex]
x = 9.51cm, -12.51cm
taking the positive values:
x = 9.51 and 12.51
W=9.51cm
L =12.51cm