已知x1、x2为方程x2+3x+1=0的两实根,则x1^2+8x^2+20=__________

2024-12-17 18:21:27
推荐回答(3个)
回答1:

X1²+3X1+1=0
∴X1²+8X2+20
=-1-3X1+8X2+20
=8X2-3X1+19
X1+X2=-3 X2=-3-X1
∴原式=8×(-3-X1)-3X1+19
=-24-8X1-3X1+19
=-11X1-5
X=(-3±√5)/2
代入 原式=-11×(-3+√5)/2-5=(23-√5)/2
或原式=-11×(-3-√5)/2-5=(23+√5)/2

回答2:

x1^2=-1-3x1
x1^2+8x2+20=-3x1+8x2+19
= - 3(x1+x2)+11x2+19
=28+11x2
=28+11*(-3±√5)/2
=(23±11√5)/2

回答3:

可以先设答案为y,然后讲y移到等号左边,再通过韦达定理中的两根和,两个积是相等的,即可解答 ,给你提供一个思路,觉得比直接给你答案会更好些

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