If alpha and beta are the zeroes of p(x)=6x^2-5x+k then find the value of k.If 1/alpha+1/beta=5
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p ( x ) = 6x² - 5x + k
On comparing with ax² + bx + c, we get
a = 6, b = - 5, c = k
Let the zeroes be α and β.
Sum of zeroes, α + β = - b/a
→ α + β = - ( - 5 )/6 = 5/6
Product of zeroes, α*β = c/a
→ α*β = k/6
Also,
(1/α) + (1/β) = 5
(α + β)/α*β = 5
Putting values, we get
→ (5/6) / (k/6) = 5
On solving this, we get
→ k = 1
Hence, the value of k is 1.
On comparing with ax² + bx + c, we get
a = 6, b = - 5, c = k
Let the zeroes be α and β.
Sum of zeroes, α + β = - b/a
→ α + β = - ( - 5 )/6 = 5/6
Product of zeroes, α*β = c/a
→ α*β = k/6
Also,
(1/α) + (1/β) = 5
(α + β)/α*β = 5
Putting values, we get
→ (5/6) / (k/6) = 5
On solving this, we get
→ k = 1
Hence, the value of k is 1.
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