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第 5 讲 解一元二次方程——公式法(一)
题一: 题面:解 方程 3x2 5x 1 0
题二: 题面:解方程 x2 x 1 0
题三: 解下列方程.
(1)
(2)2x2 2x+3=0
题四: 解下列方程.
(1)
(2)5x2 7x 3=0
− + =
− − =
2 33 04x x− + =
−
2 55 04x x+ + =
+ +2
第 5 讲 解一元二次方程——公式法(一)
题一: 见详解.
详解:∵a 3,b 5,c 1,
∴ △ b 2 4ac 13,
∴x ,
∴x1 ,x2 .
题二: 见详解.
详解:∵a 1,b 1,c 1,
∴ △ b 2 4 ac 5,
∴x ,
∴x1 ,x2 .
题三: 见详解.
详解:(1)∵a 1,b ,c ,
∴△ b2 4ac 0,
∴x ,
∴x1 x2 ;
(2)∵a 2,b 2,c 3,
∴△ b2 4ac 20 < 0,
∴原方程无实数解.
题四: 见详解.
详解:(1)∵a 1,b ,c ,
∴△ b2 4ac 0,
∴x ,
∴x1 x2 ;
= = − =
= − =
=
2 4
2
b b ac
a
− ± − = 5 13
6
±
= 5 13
6
+ = 5 13
6
−
= = − = −
= − =
=
2 4
2
b b ac
a
− ± − = 1 5
2
±
= 1 5
2
+ = 1 5
2
−
= = 3− = 3
4
= − =
=
2 4
2
b b ac
a
− ± − = 3
2
= = 3
2
= = − =
= − = −
= = 5 = 5
4
= − =
=
2 4
2
b b ac
a
− ± − = 5
2
−
= = 5
2
−3
(2)∵a 5,b 7,c 3,
∴△ b2 4ac 11 < 0,
∴原方程无实数解.
= = =
= − = −