Question

i need help in this problem:

Answer 1

\[\sqrt{x}+\sqrt{a-x}=\sqrt{y}-\sqrt{a-y}=\sqrt{z}-\sqrt{a-z}\]
prove that (y-z)(z-x)(x-y)=0

Answer 2

Is this top equation right? should it be sqrt(x) MINUS sqrt(a-x)?

Answer 3

oh yeah, sorry

Answer 4

Square them up
\[a - 2 *\sqrt{x} * \sqrt{a-x} = a - 2*\sqrt{y}*\sqrt{a-y} = a - 2*\sqrt{z}*\sqrt{a-z}\]
\[\sqrt{x} * \sqrt{a-x}=\sqrt{y}*\sqrt{a-y}=\sqrt{z}*\sqrt{a-z}\]

Answer 5

next??

Answer 6

\[x(a-x)=y(a-y)=z(a-z)\]

Answer 7

then??

Answer 8

Now they all are equal right! so I think any two of them should be same I mean either x=y or y=z or x=z

Answer 9

(y-z)(z-x)(x-y)=0

Answer 10

squaring each expression
x - 2sqrt(a-x) + a - x = -2sqrt(a-x) + a
also we get
-2sqrt(a-y) + a
- 2sqrt(a-z) + a

these 3 are equal so (a-y)=(a-x)=(a-z)
so x=y=z
there fore
(y-z)(z-x)(x-y)=0

Answer 11

@jimmy, you've dropped sqrt(x) etc.

Answer 12

there should not b sqrt{x(a-x)}??

Answer 13

@jamesj, thats my point

Answer 14

\[x(a-x)=y(a-y)=z(a-z)\]
\[xa - x^2 = ya - y^2 \]
\[ay -y^2 = az -z^2\]
\[xa - x^2 = az -z^2 \]
x,y,z can have two values but not three so eventually for the equality two variables must be same

Answer 15

or if you want to solve it more you will see
x +y = a
z+y=a
x+z=a

Answer 16

yes - i see - right james - should be 2[sqrtx* sqrt(a-x)] etc

Answer 17

Starting with Ishaan's
x(a−x)=y(a−y)
we have
x^2 - ax + y(a-y) = 0
=>\[x = \frac{1}{2} \left( a \pm \sqrt{a^2 -4ya+4y^2} \right) = \frac{1}{2} (a \pm (a-2y) ) =y \ \ or \ \ a\]

Answer 18

If x = y we're finished as x - y = 0
If x = a then y = 0 or y = a
If y = 0 and x = a, then z = 0 or a and we're finished as (x-y) or
(y-z) = 0
If y = a and x = a, then x - y = 0

Answer 19

Right!

Answer 20

thanx, guys

Answer 21

Sorry x =y or a - y.

Now modify my next argument, but it still works.

Answer 22

Or, (to beat a dead horse), starting with
x +y = a
z+y=a
x+z=a
subtract equations: (x+y)-(z+y)= 0 or (x-z)=0 and so on to get all 3 terms