The roots of the quadratic equation x2+kx+18=0x^2 + kx + 18 = 0 are α\alpha and β\beta, where kk is a constant. Given that α=2β\alpha = 2\beta, find the possible values of kk.

Worked solution

1. 利用根与系数的关系(韦达定理)列出两根之和与两根之积的方程

对于二次方程 x2+kx+18=0x^2 + kx + 18 = 0,根据根与系数的关系:
α+β=−k\alpha + \beta = -k
αβ=18\alpha \beta = 18
已知 α=2β\alpha = 2\beta,将其代入上述两式,得:
2β+β=3β=−k— (1)2\beta + \beta = 3\beta = -k \quad \text{--- (1)}
(2β)⋅β=2β2=18— (2)(2\beta) \cdot \beta = 2\beta^2 = 18 \quad \text{--- (2)}

2. 求解方程得到 $\beta$ 的值,进而求出 $k$ 的所有可能值

由方程 (2) 解出 β\beta:
2β2=18  ⟹  β2=9  ⟹  β=3 或 β=−32\beta^2 = 18 \implies \beta^2 = 9 \implies \beta = 3 \text{ 或 } \beta = -3
由方程 (1) 可知 k=−3βk = -3\beta:

  • 当 β=3\beta = 3 时,k=−3(3)=−9k = -3(3) = -9;
  • 当 β=−3\beta = -3 时,k=−3(−3)=9k = -3(-3) = 9。

因此,kk 的可能值为 99 或 −9-9。

Answer

k=9k = 9 或 k=−9k = -9(即 k=±9k = \pm 9)

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