一元二次方程求解 ( 用公式法 )1 x²-4√2x+8=0 2 (2x-1)(x-3)=43 4y²-(√2+8)y+√2=04 √2x²-√3x-√2=0 5 (y-2)(y+1)+y(y-1)=06 x²+2mx-3nx-3m²-mn+2n²=0````` 有人回答我么 ````
2019-05-27
一元二次方程求解 ( 用公式法 )
1 x²-4√2x+8=0 2 (2x-1)(x-3)=4
3 4y²-(√2+8)y+√2=0
4 √2x²-√3x-√2=0 5 (y-2)(y+1)+y(y-1)=0
6 x²+2mx-3nx-3m²-mn+2n²=0
````` 有人回答我么 ````
优质解答
x²-4√2x+8=0
a=1,b=-4√2,c=8
△=b²-4ac
=(-4√2)²-4×1×8
=32-32
=0
方程有两个相等的实数根
x1=x2=(-b±√△)/(2a)
=(4√2)/2
=2√2
(2x-1)(x-3)=4 化成一般式
2x²-7x+3=4
2x²-7x-1=0
a=2,b=-7,c=-1
△=b²-4ac
=(-7)²-4×2×(-1)
=49+8
=57
x1=(-b+√△)/(2a)
=(7+√57)/4
x2=(-b-√△)/(2a)
=(7-√57)/4
4y²-(√2+8)y+√2=0
a=4,b=-(√2+8),c=√2
△=b²-4ac
=[-(√2+8)]²-4×4×√2
=2+16√2+64-16√2
=66
y1=(-b+√△)/(2a)
=(√2+8+√66)/8
y2=(-b-√△)/(2a)
=(√2+8-√66)/8
√2x²-√3x-√2=0
a=√2,b=-√3,c=-√2
△=b²-4ac
=(-√3)²-4×√2×(-√2)
=3+8
=11
x1=(-b+√△)/(2a)
=(√3+√11)/(2√2)
=√2(√3+√11)/(2√2×√2)
=(√6+√22)/4
x2=(-b-√△)/(2a)
=(√3-√11)/(2√2)
=(√6-√22)/4
(y-2)(y+1)+y(y-1)=0 化成一般式
y²-y-2+y²-y=0
2y²-2y-2=0 方程两边同时乘1/2
y²-y-1=0
a=1,b=-1,c=-1
△=b²-4ac
=(-1)²-4×1×(-1)
=1+4
=5
y1=(-b+√△)/(2a)
=(1+√5)/2
y2=(-b-√△)/(2a)
=(1-√5)/2
x²+2mx-3nx-3m²-mn+2n²=0
x²+(2m-3n)x-3m²-mn+2n²=0
a=1,b=2m-3n ,c=-3m²-mn+2n²
△=b²-4ac
=(2m-3n)²-4×1×(-3m²-mn+2n²)
=4m²-12mn+9n²+12m²+4mn-8n²
=16m²-8mn+n²
=(4m-n)²
△=(4m-n)²≥0
√△=|4mn-n|
当△﹥0时,4m-n≠0,m≠n/4,方程有两个不相等的实数根
x1=(-b+√△)/(2a)
=(-2m+3n+|4mn-n|)/2
x2=(-b-√△)/(2a)
=(-2m+3n-|4mn-n|)/2
当△=0时,4m-n=0,m=n/4,方程有两个相等的实数根
x1=x2=(-b±√△)/(2a)
=(-2m+3n)/2
x²-4√2x+8=0
a=1,b=-4√2,c=8
△=b²-4ac
=(-4√2)²-4×1×8
=32-32
=0
方程有两个相等的实数根
x1=x2=(-b±√△)/(2a)
=(4√2)/2
=2√2
(2x-1)(x-3)=4 化成一般式
2x²-7x+3=4
2x²-7x-1=0
a=2,b=-7,c=-1
△=b²-4ac
=(-7)²-4×2×(-1)
=49+8
=57
x1=(-b+√△)/(2a)
=(7+√57)/4
x2=(-b-√△)/(2a)
=(7-√57)/4
4y²-(√2+8)y+√2=0
a=4,b=-(√2+8),c=√2
△=b²-4ac
=[-(√2+8)]²-4×4×√2
=2+16√2+64-16√2
=66
y1=(-b+√△)/(2a)
=(√2+8+√66)/8
y2=(-b-√△)/(2a)
=(√2+8-√66)/8
√2x²-√3x-√2=0
a=√2,b=-√3,c=-√2
△=b²-4ac
=(-√3)²-4×√2×(-√2)
=3+8
=11
x1=(-b+√△)/(2a)
=(√3+√11)/(2√2)
=√2(√3+√11)/(2√2×√2)
=(√6+√22)/4
x2=(-b-√△)/(2a)
=(√3-√11)/(2√2)
=(√6-√22)/4
(y-2)(y+1)+y(y-1)=0 化成一般式
y²-y-2+y²-y=0
2y²-2y-2=0 方程两边同时乘1/2
y²-y-1=0
a=1,b=-1,c=-1
△=b²-4ac
=(-1)²-4×1×(-1)
=1+4
=5
y1=(-b+√△)/(2a)
=(1+√5)/2
y2=(-b-√△)/(2a)
=(1-√5)/2
x²+2mx-3nx-3m²-mn+2n²=0
x²+(2m-3n)x-3m²-mn+2n²=0
a=1,b=2m-3n ,c=-3m²-mn+2n²
△=b²-4ac
=(2m-3n)²-4×1×(-3m²-mn+2n²)
=4m²-12mn+9n²+12m²+4mn-8n²
=16m²-8mn+n²
=(4m-n)²
△=(4m-n)²≥0
√△=|4mn-n|
当△﹥0时,4m-n≠0,m≠n/4,方程有两个不相等的实数根
x1=(-b+√△)/(2a)
=(-2m+3n+|4mn-n|)/2
x2=(-b-√△)/(2a)
=(-2m+3n-|4mn-n|)/2
当△=0时,4m-n=0,m=n/4,方程有两个相等的实数根
x1=x2=(-b±√△)/(2a)
=(-2m+3n)/2