Amaths quadratic equation

First of all, rewrite your problem as:
(x+9)^(1/2) = x - 11

If x = 7 is also the answer, then the value on the right of the equation will be -4 which is impossible in this case since:

For all positive values' square root operation, the output value should be positive too and hence x = 7 should be rejected.

I am quite sure that you are using the method of squaring both sides to solve this problem, right?
In using this method, always bear in mind that you must check the correctness of your answer(s) by directly substituting them to the original equation since such method has an appreciably high chance to introduce a false root.

Moreover, the answer should be 16 and 7 (rejected). Hope this can help you.

(x+9)^(1/2) >=0
Then, x= (x+9)^(1/2) + 11 >=11
Hence, x=7 is rejected

由於計算時會計出1正根,1負根
但代入時 25^(1/2) 不會當作 -5 , 只會=5
所以只有16這個答案

(x+9)^(1/2) + 11=x
(x+9)^(1/2) = x - 11
(x+9) = (x - 11)^2
x + 9 = x^2 - 22*x + 121
x^2 - 23 * x + 112 = 0
(x - 16) (x - 7) = 0
x = 16 or x = 7

So the answer should be x = 16 or x = 7.

But you say that the answer is x = 11 with x = 7 rejected. Are you sure???

If you have not made any typing mistake, then I think the answer should be
x = 16 or x = 7.

If the answer is really x = 11, then I think you have incorrectly typed the question.
Also, there must be some constaint NOT mentioned yet which makes the answer x = 7 become invalid. Like x > 10 or something else... What is it?? I think this question may be related to an daily live application problem instead of pure calculation. You should state the question more clearly...

Solution:

(x+9)^(1/2) + 11=x

(x+9)^(1/2)=x-11

x+9=(x-11)^2

x+9=x^2-22x+121

x^2-23x+112=0

(x-16)(x-7)=0

x=16 or x=7

Checking:

when x=16
L.H.S=(16+9)^1/2+11=16
R.H.S=16
L.H.S=R.H.S

when x=7
L.H.S=(7+9)^1/2+11=15
R.H.S=7
L.H.S≠R.H.S
∴x=16

2006-12-22 17:22:37 補充：
Ths answer must not be 11.You can check it by sub. 11 into the equation.

2006-12-22 17:27:38 補充：
Do you know why we have to do checking?

2006-12-22 17:28:06 補充：
In this case, the index=√, √ has a limitation in the equation that the answer must not be a negative number(After √ a number, the answer is positive instead of negative).When you remove the √ by square, the limitation also be removed.Therefore, we have to do checking.