?(x-3) - ?x = 3
?(x-3) = ?x + 3
square both sides:
x-3 = x + 6?x + 9
-12 = 6?x
?x = -2
by definition of ? , this has no solution.
proceeding anyway ...
square both sides again:
x = 4
since we squared, all answers must be verified:
if x = 4,
LS = ?1 - ?4
= 1 - 2
= -1
? RS
There is no solution.
verification:
http://www.wolframalpha.com/input/?i=solve+%E2%88%9A(x-3)+-+%E2%88%9Ax+%3D+3
square root (x-3)-square root(x)=3
can someone please solve this for me
2 answers
Look at it on a number line.
for positive x-3
x must be to the right of 3
but
then sqrt x must be greater than sqrt (x-3)
which is not possible.
If it were -3 on the right, that would be different
(x-3)^.5 - x^.5 = 3
square both sides
[(x-3)^.5-x^.5][(x-3)^.5-x^.5] = 9
(x-3)-2(x^.5(x-3)^.5)+x = 9
2x -2(x^.5(x-3)^.5)=12
(x^2-3x)^.5 = 6-x
x^2-3x = 36 -12x + x^2
9 x = 36
x = 4
but obviously
1 - 2 is not 3
for positive x-3
x must be to the right of 3
but
then sqrt x must be greater than sqrt (x-3)
which is not possible.
If it were -3 on the right, that would be different
(x-3)^.5 - x^.5 = 3
square both sides
[(x-3)^.5-x^.5][(x-3)^.5-x^.5] = 9
(x-3)-2(x^.5(x-3)^.5)+x = 9
2x -2(x^.5(x-3)^.5)=12
(x^2-3x)^.5 = 6-x
x^2-3x = 36 -12x + x^2
9 x = 36
x = 4
but obviously
1 - 2 is not 3