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Find the maximum of a solution expression of a quadratic equation
 Question:

Given: x_{1} and x_{2} are two solutions of the quadratic equation x^{2}  (k  1)x + k^{2} + 2k + 1 = 0.
Find the maximum of x_{1}^{2} + x_{2}^{2}
 Solution:
 If the quadratic equation has two solutions, then b^{2}  4ac >= 0
 For this quadratic equation: a = 1, b =  (k  1), c = k^{2} + 2k + 1
 b^{2}  4ac = [(k  1)]^{2}  4(1)(k^{2} + 2k + 1)
 = (1)^{2} (k  1)^{2}  4(k^{2} + 2k + 1)
 = k^{2}  2k + 1  4k^{2}  8k  4
 = 3k^{2}  10k  3
 = (3k^{2} + 10k + 3)
 Since the quadratic equation has two solutions, so b^{2}  4ac >= 0
 then (3k^{2} + 10k + 3) >= 0
 Both side of the inequality multiply (1) and the inequality change its direction
 3k^{2} + 10k + 3 <= 0
 k_{1} <= [10 + Sqrt (10^{2}  4 × 3 × 3)]/(2 × 3)
 = [10 + Sqrt (100  36)]/6
 = [10 + Sqrt (64)]/6
 = [10 + 8]/6
 = 2/6
 = 1/3
 k_{2} <= [10  Sqrt (10^{2}  4 × 3 × 3)]/(2 × 3)
 = [10  Sqrt (100  36)]/6
 = [10  Sqrt (64)]/6
 = [10  8]/6
 = 18/6
 = 3
 so the k is in the range of 3 <= k <= 1/3
 x_{1}^{2} + x_{2}^{2}
 = (x_{1} + x_{2})^{2}  2x_{1}x_{2}
 = (k  1)^{2}  2(k^{2} + 2k + 1)
 = k^{2}  2k + 1  2k^{2}  4k 2
 =  k^{2}  6k  1
 =  (k^{2} + 6k + 1)
 =  (k^{2} + 6k + 3^{2}  3^{2} + 1)
 =  (k^{2} + 6k + 3^{2}) + 9  1
 =  (k + 3)^{2} + 8
 When k = 3, x_{1}^{2} + x_{2}^{2} = 8
 k = 3 is in the range of k. So when k = 3, x_{1}^{2} + x_{2}^{2} reach the maximum 8.
Rule used: If quadratic equation x^{2} + px + q has two solution x_{1} and x_{2}, then the sum of these two solution is p. That is,
x_{1} + x_{2} = p. The product of these two solutions is q. That is, x_{1} x_{2} = q.