SOLUTION: find the range of values for p such that 3X^2 + 3px + p^2 = 1 has real roots
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Question 71000: find the range of values for p such that 3X^2 + 3px + p^2 = 1 has real roots
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
find the range of values for p such that
3X^2 + 3px + p^2 = 1 has real roots
3X^2+3PX+(P^2-1)=0
DISCRIMINANT = (3P)^2-4(3)(P^2-1)=9P^2-12P^2+12= 12-3P^2
HENCE FOR REAL ROOTS
12-3P^2>=0
4-P^2>=0
4>=P^2
2>=|P|
|P|<=2
RANGE OF P IS [-2,2]
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