Question

If the roots of equation xsquare minus px plus one =zero is real ,then find 'p'

Answer 1

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 x²-px+1=0
Roots are real if D≥0
D=b²-4ac
(-p)²-4(1)(1)≥0
p²-4≥0
p²≥4
p≥√4
p≥±2
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