By completing the square find in terms of k the solutions of the equation x^2 - 4kx + 6 =0?
回答 (2)
2020-09-23 2:10 pm
x² - 4kx + 6 = 0
x² - 4kx = -6
x² - 2(2k)x + (2k)² = -6 + (2k)²
(x - 2k)² = 4k² - 6
x - 2k = ± √(4k² - 6)
x = 2k ± √(4k² - 6)
x² - 4kx = -6
x² - 2(2k)x + (2k)² = -6 + (2k)²
(x - 2k)² = 4k² - 6
x - 2k = ± √(4k² - 6)
x = 2k ± √(4k² - 6)
2020-09-23 5:00 pm
x² - 4kx + 6 = 0
x² - 4kx = - 6
x² - 4kx + (2k)² - (2k)² = - 6
x² - 4kx + (2k)² = (2k)² - 6
(x - 2k)² = 4k² - 6
x - 2k = ± √(4k² - 6)
x = 2k ± √(4k² - 6)
x₁ = 2k + √(4k² - 6)
x₂ = 2k - √(4k² - 6)
x² - 4kx = - 6
x² - 4kx + (2k)² - (2k)² = - 6
x² - 4kx + (2k)² = (2k)² - 6
(x - 2k)² = 4k² - 6
x - 2k = ± √(4k² - 6)
x = 2k ± √(4k² - 6)
x₁ = 2k + √(4k² - 6)
x₂ = 2k - √(4k² - 6)
收錄日期: 2021-05-01 09:39:36
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