2題一元二次方程(15點)

如果我誤解了你所指的『整個 √』，請再提出你的問題。


1. √(2-7x)-√(x+3)=3
√(2-7x)=3+√(x+3)

Squaring both sides:
2-7x=9+6√(x+3)+(x+3)
-10-8x=6√(x+3)
-5-4x=3√(x+3)

Squaring both sides:
25+40x+16x^2=9x+27
16x^2+31x-2=0
(16x-1)(x+2)=0
16x-1=0 or x+2=0
x=1/16 or x=-2

Checking:
When x=1/16
L.H.S.=√(2-7/16)-√(1/16+3)
=√(25/16)-√(49/16)
=5/4-7/4
=-1/2
≠R.H.S.

When x=-2
L.H.S.=√(2+14)-√(-2+3)
=√16-√1
=4-1
=3
=R.H.S.

∴ x=-2


2. √(x+5)+√(2x+3)=√(8-x) [相信右邊應該不是v]

Squaring both sides:
(x+5)+2√[(x+5)(2x+3)]+(2x+3)=8-x
2√[(x+5)(2x+3)]=-4x
√[(x+5)(2x+3)]=-2x

Squaring both sides:
(x+5)(2x+3)=4x^2
2x^2+13x+15=4x^2
2x^2-13x-15=0
(2x-15)(x+1)=0
x=15/2 or x=-1

Checking:
When x=15/2
L.H.S.=√(15/2+5)+√(15+3)
=√(25/2)+√18
=√(50/4)+3√2
=(5/2)√2+3√2
=(11/2)√2
R.H.S.=√(8-15/2)
=√(1/2)
=2√2
L.H.S.≠R.H.S.

When x=-1
L.H.S.=√(-1+5)+√(-2+3)
=√4+√1
=2+1
=3
R.H.S.=√(8+1)
=√9
=3
L.H.S.=R.H.S.

∴ x=-1

抱歉真的不太明白你的意思~~