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Let k be the largest integer such that the equation (x−1)2 + 2kx + 11 = 0 has no real roots. If y is a positive real number, then the least possible value of k/4y + 9y is
Correct answer is '6'. Can you explain this answer?
Most Upvoted Answer
Let kbe the largest integer such that the equation(x−1)2 + 2kx +...

Since the equation above has no real roots, the discriminant of the equation should be negative.

The largest integral value that k can take is 4.
Now, we need to minimize k/4y + 9y where k takes the largest integral value and y is positive…

1/y & 9y are both positive real numbers, therefore, their A.M is greater than equal to their G.M.
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Community Answer
Let kbe the largest integer such that the equation(x−1)2 + 2kx +...
Explanation:

Finding the value of k:
- Given equation: (x-1)^2 + 2kx + 11 = 0
- For no real roots, discriminant < />
- Discriminant = 4k^2 - 4(11-(1)^2) < />
- 4k^2 - 40 < />
- k^2 < />
- k = 3 (largest integer)

Finding the least possible value of k/4y + 9y:
- Given y is a positive real number
- k = 3, y > 0
- k/4y + 9y = 3/(4y) + 9y
- To minimize the value, make the first term as small as possible
- Minimize 3/(4y) by maximizing y
- Let y = 1/4, then 3/(4y) = 3
- Substitute y = 1/4 in the expression k/4y + 9y
- 3/1 + 9(1/4) = 3 + 9/4 = 12/4 + 9/4 = 21/4 = 5.25
Therefore, the least possible value of k/4y + 9y is 5.25, not 6.
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Let kbe the largest integer such that the equation(x−1)2 + 2kx + 11 = 0has no real roots. If y is a positive real number,then the least possible value of k/4y + 9y isCorrect answer is '6'. Can you explain this answer?
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