If alpha and beta are zeroes of 2x2+5x+k such that alpha2+beta2 + alpha beta = 21/4
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Question should be:
- If alpha and beta are zeroes of 2x²+5x+k such that (alpha)²+(beta)² + (alpha)(beta) = 21/4
ANSWER:
- Value of k is 2
GIVEN:
- P(x) = 2x²+5x+k
- α²+β²+αβ = 21/4
TO FIND:
- Value of 'k'.
SOLUTION:
P(x) = 2x²+5x+k
Here:
Coefficient of x² = 2
Coefficient of x = 5
Constant term = k
Formula:
=> Sum of zeroes (α+β) = -(Coefficient of x)/Coefficient of x²
=> Product of zeroes (αβ) = Constant term/ Coefficient of x²
=> Sum of zeros = -5/2
=> Product of zeros = k/2
Now:
Value of k is 2.
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