Question on: WAEC Mathematics - 2010
If the sum of the roots of the equation (x - p)(2x - 1) - 0 is 1, find the value of x
A
1\(\frac{1}{2}\)
B
\(\frac{1}{2}\)
C
-\(\frac{3}{2}\)
D
-1\(\frac{1}{2}\)
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Correct Option: A
(x - p)(2x + 1) = 0
2x2 + x - 2px - p = 0
2x2 + x (1 - 2p) - p = 0
2x2 - (2p - 1)x - p = 0
divide through by 2
x2 - \(\frac{(2p - 1)}{2}\)x - \(\frac{p}{2}\) = 0
compare to x2 - (sum of roots)x + product of roots = 0
sum of roots = \(\frac{2p - 1}{2}\)
But sum of roots = 1
Given; \(\frac{2p - 1}{2}\) = 1
2p - 1 = 2 x 1
2p - 1 = 2
2p = 2 + 1 = 3
p = \(\frac{3}{2}\)
p = 1\(\frac{1}{2}\)
2x2 + x - 2px - p = 0
2x2 + x (1 - 2p) - p = 0
2x2 - (2p - 1)x - p = 0
divide through by 2
x2 - \(\frac{(2p - 1)}{2}\)x - \(\frac{p}{2}\) = 0
compare to x2 - (sum of roots)x + product of roots = 0
sum of roots = \(\frac{2p - 1}{2}\)
But sum of roots = 1
Given; \(\frac{2p - 1}{2}\) = 1
2p - 1 = 2 x 1
2p - 1 = 2
2p = 2 + 1 = 3
p = \(\frac{3}{2}\)
p = 1\(\frac{1}{2}\)
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