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Sagot :
The area of a rectangle is length L times width W.
Area = L * W.
We are told that L = W + 25.
So if the area = L * W
We can substitute L with W + 25 to give us
6984 = L * W = (W + 25 ) * W = W^2 + 25W.
W^2 + 25W = 6984. We can turn this into a quadratic equation by removing 6984 from both sides
W^2 + 25W - 6984 = 0
To solve a quadratic ax^x + bx + c = 0 we use this formula
[tex](-b+/- \sqrt{-b^{2} -4ac})/2a [/tex]
In our quadratic a = 1, b=25, c=-6984
-b = -25
b2 - 4ac = 25*25 - 4 * 1 * -6984 = 625 + 27936 = 28561
SQRT(28561) = 169
So we have two answers (from the +/- in the quadratic solver)
(-b+169)/(2*1) = (-25+169)/2 = 144/2 = 72or(-b-169)/(2*1) = (-194)/2 = -97Since we can't have a negative length, 72 must be the value for W
If W is 72 then L = W + 25 = 97
We can now test this by saying A = L * W = 97 * 72 = 6984, the original area given.
Area = L * W.
We are told that L = W + 25.
So if the area = L * W
We can substitute L with W + 25 to give us
6984 = L * W = (W + 25 ) * W = W^2 + 25W.
W^2 + 25W = 6984. We can turn this into a quadratic equation by removing 6984 from both sides
W^2 + 25W - 6984 = 0
To solve a quadratic ax^x + bx + c = 0 we use this formula
[tex](-b+/- \sqrt{-b^{2} -4ac})/2a [/tex]
In our quadratic a = 1, b=25, c=-6984
-b = -25
b2 - 4ac = 25*25 - 4 * 1 * -6984 = 625 + 27936 = 28561
SQRT(28561) = 169
So we have two answers (from the +/- in the quadratic solver)
(-b+169)/(2*1) = (-25+169)/2 = 144/2 = 72or(-b-169)/(2*1) = (-194)/2 = -97Since we can't have a negative length, 72 must be the value for W
If W is 72 then L = W + 25 = 97
We can now test this by saying A = L * W = 97 * 72 = 6984, the original area given.
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