SOLUTION: The length of a rectangle is x and the width is sqrt(3x + 1) If the width is 1 less than the length, what are the dimensions of the rectangle?

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Question 363533: The length of a rectangle is x and the width is sqrt(3x + 1) If the width is 1 less than the length, what are the dimensions of the rectangle?

Answer by vasumathi(46) (Show Source):
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The length of a rectangle is x and the width is sqrt(3x + 1) If the width is 1 less than the length, what are the dimensions of the rectangle?
Solution:length = x
width = sqrt(3x+1)
since the width is one less than the length we have...
sqrt(3x+1) = x-1
let us square on both sides
sqrt(3x+1)^2 = (x-1)^2
3x+1 = x^2 -2x +1 ( I used FOIL.(x-1)^2 = (x-1)*(x-1) = x^2-x-x+1=x^2-2x+1)
subtract 3x +1 on both sides'
3x+1-3x-1 = x^2-2x+1-3x-1
x^2-5x=0
x(x-5)= 0
so x= 0 and x = 5 these are the two values of x
That is the length = x = 5 (since we cant have x = 0 as length cant be 0)
The width = sqrt (3x+1)= sqrt (3*5 +1) = sqrt (15+1) = sqrt (16) = 4
These are the dimensions