For what values of p will the equation
(1+p)x^2+2(1+2p)x+(1+p)=0 has coincident roots?
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Answer:
The roots are coincident when b²-4ac = 0
b = 2(1+2p)
= 2 + 4p
a = 1+ p
c = 1+ p
Putting values in equation
(2+4p)² - 4(1+p)(1+p) =0
4 + 16p² + 16p - 4(1 + p² + 2p) = 0
4 + 16p² + 16p - 4 - 4p² - 8p = 0
12p² + 8p = 0
4p (3p + 2) = 0
4p=0 3p+2=0
p=0 and p=-2/3
The values of p are (0) and (-2/3).
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