韦达定理的推导公式 设方程ax^2+bx+c=0 (a≠0 且△=b^2-4ac≥0),用x来表示y1、y2
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ax^2+bx+c=0
两边同除以a
x^2 +b/a x +c/a = 0
配方
(x+ b/(2a) )^2 +c/a -b^2/(4a^2) = 0
(x+ b/(2a) )^2 =b^2/(4a^2) - c/a
开方
x+b/(2a) = +或- √[b^2/(4a^2) - c/a ]
y1 = -b/(2a) + √[b^2/(4a^2) - c/a ] = [-b + √(b^2-4ac)] /(2a)
y2 = -b/(2a) - √[b^2/(4a^2) - c/a ] = [-b - √(b^2-4ac)] /(2a)
两边同除以a
x^2 +b/a x +c/a = 0
配方
(x+ b/(2a) )^2 +c/a -b^2/(4a^2) = 0
(x+ b/(2a) )^2 =b^2/(4a^2) - c/a
开方
x+b/(2a) = +或- √[b^2/(4a^2) - c/a ]
y1 = -b/(2a) + √[b^2/(4a^2) - c/a ] = [-b + √(b^2-4ac)] /(2a)
y2 = -b/(2a) - √[b^2/(4a^2) - c/a ] = [-b - √(b^2-4ac)] /(2a)
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