Question

解下列一元二次方程 x^-49=0 2y^2=128 2x^-1/2=0 (3y-7)^2=3 7-2x^2=-15 t^2-45=0 (x+5)^2=16

Answer 1

答：

x^-49=0
x²=49
x²=7²
x=7或者x=-7

2y^2=128
y²=64
y²=8²
y=8或者y=-8

2x^-1/2=0
2x²=1/2
x²=1/4
x²=(1/2)²
x=1/2或者x=-1/2

(3y-7)^2=3
3y-7=√3或者3y-7=-√3
y=(7+√3)/3或者y=(7-√3)/3

7-2x^2=-15
2x²=15+7
2x²=22
x²=11
x=√11或者x=-√11

t^2-45=0
t²=45
t=√45或者t=-√45
所以：t=3√5或者t=-3√5

(x+5)^2=16
x+5=-4或者x+5=4
所以：x=-9或者x=-1

Answer 2

x^-49=0
x=±7
2y^2=128
y=±8
2x^-1/2=0
y=±1/2
(3y-7)^2=3
y=(7±√3)/3
7-2x^2=-15
x=±√42/2
t^2-45=0
t=±3√5
(x+5)^2=16
x=-5±4

追问

x1和x2在哪里