if a and b are the roots of the quadratic equation 3x^2+6x-2=0.find the values of the following expressions.
(A)a^2+B^2
(B)A/B+B/A
(C)2/a^2+2/b^2
f.4一元二次方程(II)(5分)
2014-10-10 5:50 am
回答 (2)
2014-10-10 7:10 am
✔ 最佳答案
a+b = -6/3 = -2a*b = -2/3
A)
a^2 + b^2 = (a+b)^2 -2ab = -2 +4/3 = -2 / 3
B)
A/B + B/A
=( A^2 + B^2 ) / AB = ( -2/3 ) / ( -2/3 ) = 1
C)
2/a^2 + 2/b^2
=(2b^2+2a^2) / (a^2*b^2)
=(-4/3) / (-4/9)
=3
參考: ME
2014-10-10 7:31 am
(A)
αand β are the roots of 3x² + 6x - 2 = 0
Sum of roots : α + β = -6/3 = -2
Product of roots : αβ = -2/3
α² + β²
= (α² + 2αβ + β²) -2αβ
= (α + β)² - 2αβ
= (-2)² - 2(-2/3)
= 16/3
(B)
(α/β) + (β/α)
= (α²/αβ)+ (β²/αβ)
= (α² + β²)/αβ
= (16/3)/(-2/3)
= -8
2014-10-09 23:32:06 補充：
(C)
(2/α²) + (2/β²)
= (2β²/α²β²) + (2α²/α²β²)
= 2(α² + β²)/(αβ)²
= 2(16/3)/(-2/3)²
= 24
αand β are the roots of 3x² + 6x - 2 = 0
Sum of roots : α + β = -6/3 = -2
Product of roots : αβ = -2/3
α² + β²
= (α² + 2αβ + β²) -2αβ
= (α + β)² - 2αβ
= (-2)² - 2(-2/3)
= 16/3
(B)
(α/β) + (β/α)
= (α²/αβ)+ (β²/αβ)
= (α² + β²)/αβ
= (16/3)/(-2/3)
= -8
2014-10-09 23:32:06 補充：
(C)
(2/α²) + (2/β²)
= (2β²/α²β²) + (2α²/α²β²)
= 2(α² + β²)/(αβ)²
= 2(16/3)/(-2/3)²
= 24
收錄日期: 2021-04-15 16:47:34
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