a.maths -- quadratic equation (very urgent~~)
2007-09-17 12:27 am
1. If a is the positive root of the equation x²+3x-1=0. Find the value of a^4+4a^3-3a²-2a+1 (leave your answer in surd form) (希望大家幫幫忙, 計左好耐~)
回答 (2)
2007-09-17 12:50 am
✔ 最佳答案
x^2 + 3x-1=0positive root, x = (-3+(3^2-4(1)(-1))^0.5)/2 = (-3+(13)^0.5)/2
a is the root of x^2+3x-1 = 0
=> a^2+3a-1 =0
By using long division,
a^4+4a^3-3a^2 -2a+1= (a^2+3a-1)(a^2+a-5) +14a+6
= 14a+6 (because a^2+3a -1 =0 )
= 14(-3+(13)^0.5)/2 +6
= -15+7(13)^0.5
參考: My calculation
2007-09-17 1:15 am
1.a^2+3a-1=0
(a+3/2)^2-9/4-1=0
(a+3/2)^2=13/4
a=開方(13/4)-3/2 or a=-開方(13/4)-3/2(rejected) where a is positive
之後你代番落去a^4+4a^3-3a²-2a+1就ok啦
(a+3/2)^2-9/4-1=0
(a+3/2)^2=13/4
a=開方(13/4)-3/2 or a=-開方(13/4)-3/2(rejected) where a is positive
之後你代番落去a^4+4a^3-3a²-2a+1就ok啦
收錄日期: 2021-04-13 13:31:41
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