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標題:
a.maths -- quadratic equation (very urgent~~)
發問:
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~)
最佳解答:
x^2 + 3x-1=0 positive 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
其他解答:
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.maths -- quadratic equation (very urgent~~)
發問:
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~)
最佳解答:
x^2 + 3x-1=0 positive 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
其他解答:
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
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