Question

墨萱( ˶ ❛ ꁞ ❛ ˶ )

求解❗❗

5 解下列各一元二次方程式:
(1) (x+3)² = 4 (x+3)
P.163 15 5
(x-2)(2x+5)-(2-x)(5x+2)=0

Answer 1

Mer

令x+3=y
y^2=4y ==>y^2-4y=0
y(y-4)=0 ==>y=0或4
故x+3=0或4 ==>x=-3或1

墨萱( ˶ ❛ ꁞ ❛ ˶ )

謝謝

Mer

2-x=-(x-2)
原式:(x-2)(2x+5)+(x-2)(5x+2)=0
(x-2)[(2x+5)+(5x+2)]=0
(x-2)(7x+7)=0 ==>x=2或-1