*Solve. √8-2x = x?

2009-10-20 3:19 pm
a. {2}
b. {-4}
c. {2,-4}
d. Ø

回答 (8)

2009-10-23 6:20 am
✔ 最佳答案
√(8 - 2x) = x
8 - 2x = x²
x² + 2x - 8 = 0
= (x + 4)(x - 2)

Answer: a. {2}, x = - 4 or x = 2
2009-10-20 4:56 pm
√8 - 2x = x
3x = √8
x = √8 / 3
x = 2√2 / 3
2009-10-20 3:36 pm
√(8 - 2x) = x
8 - 2x = x^2
x^2 + 2x - 8 = 0
x^2 + 4x - 2x - 8 = 0
(x^2 + 4x) - (2x + 8) = 0
x(x + 4) - 2(x + 4) = 0
(x + 4)(x - 2) = 0

x + 4 = 0
x = -4

x - 2 = 0
x = 2

∴ x = -4, 2
(answer c)
2009-10-20 3:29 pm
sqrt ( 8 - 2x ) = x .......(1)

Squaring both sides

8 - 2x = x^2

x^2 + 2x - 8 = 0

( x + 4 )( x - 2 ) = 0

x = - 4 or x = 2.
.................................................................................

But putting x = - 4 in (1) gives sqrt ( 16 ) = -4 which is false.
Such a root is called an extraneous root.
.................................................................................

Putting x = 2 gives sqrt ( 4 ) = 2 which is ' legal '.
..................................................................................

Hence, there is only one correct solution, namely, a.{ 2 }........Ans.
2009-10-20 3:27 pm
Um.... none.

√8 - 2x = x

First isolate x terms by adding 2x to both sides.

√8 = 3x

√8 = 2√2 (as √8 = √(4*2) or √4 * √2

Divide both sides by 3.

(2√2)/3 = x

If d. Ø means no solutions available, then d is the right answer.
參考: If my answer is wrong, then you have to make sure to include the parantheses when you put them down, because parantheses are CRUCIAL and can not be left out.
2009-10-20 3:26 pm
Square both sides:

(√8-2x)^2 = x^2

8 - 2x = x^2

Then bring everything to one side to make a quadratic equation:

x^2 + 2x - 8 = 0

And factor:

x^2 + 2x - 8 = (x + 4)(x - 2)

So your solution set is {x | x = 2, -4}
2009-10-20 3:25 pm
√(8-2x) = x
square both sides
8-2x=x²
0=x²+2x-8
0=(x+4)(x-2)
x = -4 or 2
c {2,-4}
2009-10-20 3:24 pm
It should be written √(8-2x) = x.

Square both sides of the equation
8 - 2x = x²
x² + 2x - 8 = 0
(x-2)(x+4) = x² + 2x - 8
x = 2, -4


收錄日期: 2021-05-01 12:59:08
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091020071938AAoChaI

檢視 Wayback Machine 備份