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因式分解法解一元二次方程题,顺便告诉一下秘诀
x^2+2mx+m^2-n^2=0
x^2-ax+(a^2/4)-b^2=0
X^2-bx-2b^2=0
大哥大姐们顺便告诉我一下因式分解法解一元二次方程的秘诀吧
人气:233 ℃ 时间:2020-04-20 18:13:17
解答
m^2-n^2 = (m+n)(m-n)
x^2+2mx+m^2-n^2=0
(x+m-n)(x+m+n) = 0
x1=n-m x2=-n-m
--
(a^2/4)-b^2 = (a/2+b)(a/2-b)
x^2-ax+(a^2/4)-b^2=0
(x-a/2+b)(x-a/2-b)=0
x1=a/2+b x2=a/2-b
--
x^2-bx-2b^2=0
(x - 2b)(x + b) = 0
x1 = 2b x2 = -b
秘诀就是靠RP...
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