The sum of two positive numbers is 26 and their product is 144. The positive difference between them is:
10
Let the two positive numbers be \(a\) and \(b\) with \(a+b=26\) and \(ab=144\).
Use the identity \((a-b)^2=(a+b)^2-4ab\).
Substitute: \((a-b)^2 = 26^2 - 4\times 144 = 676 - 576 = 100\).
Therefore \(|a-b| = \sqrt{100} = 10\).
Hence, the positive difference between the two numbers is 10.
In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.
I. x2 – 26x + 165 = 0
II. y2 – 38y + 357 = 0
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(x 2- 6xy + 9y 2) - 25
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AP = PQ = QB, then the mid point of PQ is
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