The quadratic equation x2−4x+2=0x^2 - 4x + 2 = 0 has roots α\alpha and β\beta. Find a quadratic equation, with integer coefficients, whose roots are 2α2\alpha and 2β2\beta.

Worked solution

1. 由原方程利用根与系数的关系求出 $\alpha+\beta$ 与 $\alpha\beta$ 的值

已知一元二次方程 x2−4x+2=0x^2 - 4x + 2 = 0 的两根为 α\alpha 和 β\beta。
根据二次方程根与系数的关系(韦达定理):
α+β=4\alpha + \beta = 4
αβ=2\alpha \beta = 2

2. 计算新方程两根 $2\alpha$ 与 $2\beta$ 的和与积

新方程的两根为 2α2\alpha 和 2β2\beta。
新根之和为:
2α+2β=2(α+β)=2×4=82\alpha + 2\beta = 2(\alpha + \beta) = 2 \times 4 = 8
新根之积为:
(2α)(2β)=4αβ=4×2=8(2\alpha)(2\beta) = 4\alpha\beta = 4 \times 2 = 8

3. 利用新根的和与积构造出满足条件的二次方程

以 2α2\alpha 和 2β2\beta 为根的二次方程可表示为:
x2−(新根之和)x+(新根之积)=0x^2 - (\text{新根之和})x + (\text{新根之积}) = 0
代入算得的新根之和与积:
x2−8x+8=0x^2 - 8x + 8 = 0
该方程系数均为整数,符合题意。

Answer

x2−8x+8=0x^2 - 8x + 8 = 0

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