write down the roots of the equation ax^2 + bx + c = 0 .
要點做架???
F.4math
2013-10-01 11:07 pm
回答 (2)
2013-10-01 11:30 pm
✔ 最佳答案
write down the roots of the equation ax^2+bx+c=0,a<>0Sol
ax^2+bx+c=0
ax^2+bx=－c
x^2+bx/a=－c/a
x^2+2*x*[b/(2a)]=－c/a
x^2+2*x*[b/(2a)]+[b/(2a)]^2=[b/(2a)]^2－c/a
[x+b/(2a)]^2=b^2/(4a^2)－ac/a^2
[x+b/(2a)]^2=(b^2－4ac)/(4a^2)
x+b/(2a)=+/-√(b^2－4ac)/(2a)
x=[－b+/-√(b^2－4ac)]/(2a)
2013-10-01 11:30 pm
It should be done as this.
2013-10-01 15:32:40 補充：
ax^2+bx+c=0
ax^2+bx=-c
x^2+b/a x=-c/a
x^2+b/a x+(〖b/2a)〗^2=-c/a+(〖b/2a)〗^2
〖(x+b/2a)〗^2=-4ac/(4a^2 )+b^2/(4a^2 )
〖(x+b/2a)〗^2=(b^2-4ac)/(4a^2 )
x+b/2a=±√((b^2-4ac)/(4a^2 ))
x+b/2a=(±√(〖 b〗^2-4ac))/2a
x=-b/2a±√(〖 b〗^2-4ac)/2a
x=(-b±√(b^2-4ac))/2a
2013-10-01 15:32:40 補充：
ax^2+bx+c=0
ax^2+bx=-c
x^2+b/a x=-c/a
x^2+b/a x+(〖b/2a)〗^2=-c/a+(〖b/2a)〗^2
〖(x+b/2a)〗^2=-4ac/(4a^2 )+b^2/(4a^2 )
〖(x+b/2a)〗^2=(b^2-4ac)/(4a^2 )
x+b/2a=±√((b^2-4ac)/(4a^2 ))
x+b/2a=(±√(〖 b〗^2-4ac))/2a
x=-b/2a±√(〖 b〗^2-4ac)/2a
x=(-b±√(b^2-4ac))/2a
收錄日期: 2021-05-02 10:47:54
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