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求一元二次方程的根 [列明步驟]
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最佳解答:
這兩題要用根與係數的關係： 若ax^2+bx+c=0 (a≠0)的兩根為α、β，則α+β = -b/a；αβ = c/a 1.若a,b 為二次方程2x^2+4x-1=0的根,求: a)a+b = -4/2 = -2 b)ab = -1/2 =-1/2 c)1/a+1/b = (b+a)/ab = -2/(-1/2) =4 d)(a+1)(b+1) = ab+ a + b+1 = -1/2 + -2 +1 = -3/2 2.若a,b為二次方程x^2-3x-5=0的根, a+b= - (-3)/1 =3 ab = -5/1 = -5 求: a)(a-1)(b-1) =ab -a -b +1 =ab -(a+b) +1 = -5 -3 +1 = -7 b)a^2+b^2 = a^2+2ab+b^2 -2ab =(a+b)^2 -2ab =3^2 -2(-5) =9+10 =19 c)a/b+b/a =a^2/ab + b^2 /ab = (a^2+b^2)/ab =19/-5 = -19/5 d)(a+1/a)(b+1/b) = ab + (a/b +b/a) +1/ab = -5+ (-19/5) +1/(-5) = -9
其他解答:
1.若a,b 為二次方程2x^2+4x-1=0的根,求: Method1: You should know that sum of the roots=-b/a and product of the roots=c/a a) a+b=-4/2=-2 b) ab=-1/2 c) 1/a+1/b =a+b/ab =-2/(-1/2) =4 d) (a+1)(b+1) =ab+a+b+1 =-1/2-2+1 =-3/2 Method2: Use quadratic formula 2x^2+4x-1=0 x={-4+-sqrt[(4)^2-4(2)(-1)]}/4 x={-4+-sqrt24}/4 a=(-4+sqrt24)/4 b=(-4-sqrt24)/4 a)a+b=(-4+sqrt24-4-sqrt24)/4=-8/4=-2 b)ab=(16-24)/16=-1/2 c) 1/a+1/b =a+b/ab =-2/(-1/2)=4 d) (a+1)(b+1) =ab+a+b+1 =-1/2-2+1=-3/2 2.若a,b為二次方程x^2-3x-5=0的根,求: a)(a-1)(b-1) b)a^2+b^2 c)a/b+b/a d)(a+1/a)(b+1/b) ANS: a+b=3, ab=-5 a) (a-1)(b-1) =ab-a-b+1 =ab-(a+b)+1 =-5-3+1=-7 b) a^2+b^2 =(a+b)^2-2ab =(3)^2-2(-5) =9+10 =19 c)a/b+b/a =a^2+b^2/ab =-19/5 d)(a+1/a)(b+1/b) =ab+a/b+b/a+1/ab =-5-19/5-1/5 =-5-20/5 =-9
求一元二次方程的根 [列明步驟]
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此文章來自奇摩知識+如有不便請留言告知
1.若a,b 為二次方程2x^2+4x-1=0的根,求: a)a+b b)ab c)1/a+1/b d)(a+1)(b+1) 2.若a,b為二次方程x^2-3x-5=0的根,求: a)(a-1)(b-1) b)a^2+b^2 c)a/b+b/a d)(a+1/a)(b+1/b)最佳解答:
這兩題要用根與係數的關係： 若ax^2+bx+c=0 (a≠0)的兩根為α、β，則α+β = -b/a；αβ = c/a 1.若a,b 為二次方程2x^2+4x-1=0的根,求: a)a+b = -4/2 = -2 b)ab = -1/2 =-1/2 c)1/a+1/b = (b+a)/ab = -2/(-1/2) =4 d)(a+1)(b+1) = ab+ a + b+1 = -1/2 + -2 +1 = -3/2 2.若a,b為二次方程x^2-3x-5=0的根, a+b= - (-3)/1 =3 ab = -5/1 = -5 求: a)(a-1)(b-1) =ab -a -b +1 =ab -(a+b) +1 = -5 -3 +1 = -7 b)a^2+b^2 = a^2+2ab+b^2 -2ab =(a+b)^2 -2ab =3^2 -2(-5) =9+10 =19 c)a/b+b/a =a^2/ab + b^2 /ab = (a^2+b^2)/ab =19/-5 = -19/5 d)(a+1/a)(b+1/b) = ab + (a/b +b/a) +1/ab = -5+ (-19/5) +1/(-5) = -9
其他解答:
1.若a,b 為二次方程2x^2+4x-1=0的根,求: Method1: You should know that sum of the roots=-b/a and product of the roots=c/a a) a+b=-4/2=-2 b) ab=-1/2 c) 1/a+1/b =a+b/ab =-2/(-1/2) =4 d) (a+1)(b+1) =ab+a+b+1 =-1/2-2+1 =-3/2 Method2: Use quadratic formula 2x^2+4x-1=0 x={-4+-sqrt[(4)^2-4(2)(-1)]}/4 x={-4+-sqrt24}/4 a=(-4+sqrt24)/4 b=(-4-sqrt24)/4 a)a+b=(-4+sqrt24-4-sqrt24)/4=-8/4=-2 b)ab=(16-24)/16=-1/2 c) 1/a+1/b =a+b/ab =-2/(-1/2)=4 d) (a+1)(b+1) =ab+a+b+1 =-1/2-2+1=-3/2 2.若a,b為二次方程x^2-3x-5=0的根,求: a)(a-1)(b-1) b)a^2+b^2 c)a/b+b/a d)(a+1/a)(b+1/b) ANS: a+b=3, ab=-5 a) (a-1)(b-1) =ab-a-b+1 =ab-(a+b)+1 =-5-3+1=-7 b) a^2+b^2 =(a+b)^2-2ab =(3)^2-2(-5) =9+10 =19 c)a/b+b/a =a^2+b^2/ab =-19/5 d)(a+1/a)(b+1/b) =ab+a/b+b/a+1/ab =-5-19/5-1/5 =-5-20/5 =-9
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