- 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 + square root of (10^{2}- 4 × 3 × 3)]/(2 × 3) - = [-10 + square root of (100 - 36)]/6
- = [-10 + square root of 64]/6
- = [-10 + 8]/6
- = -2/6
- = -1/3
- k
_{2}<= [-10 - square root of (10^{2}- 4 × 3 × 3)]/(2 × 3) - = [-10 - square root of (100 - 36)]/6
- = [-10 - square root of 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.