Find the value of p for which the roots of quadratic equation 3x sqr - px + 3 = 0
are real where p > 0.
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Answered by
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3x^2-px+3
Here a=3 b=p c=3
Delta=b^2-4ac
=>p^2-4×3×3
=>p^2-36
=>p^2=36
=>p^2=6^2
=>p= plus or minus 6
The value of p is plus or minus6
Answered by
1
3 x² - p x + 3 = 0
Discriminant = p² - 36 must be greater than or equal to 0, for real roots.
p ≥ 6 or p ≤ - 6
P belongs to the real value set given by:
p ∈ (-∞ , -6 ] U [6, ∞)
Discriminant = p² - 36 must be greater than or equal to 0, for real roots.
p ≥ 6 or p ≤ - 6
P belongs to the real value set given by:
p ∈ (-∞ , -6 ] U [6, ∞)
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