SOLUTION: the difference between an integer and its square root is 6. find the integer.

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Question 1101935: the difference between an integer and its square root is 6. find the integer.
Found 3 solutions by Alan3354, greenestamps, ikleyn:
Answer by Alan3354(69443) About Me  (Show Source):
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


This is one example where giving the "obvious" answer without showing any work is of little help, because the solution by a formal algebraic process has some things that the student needs to watch our for.

The given information says x+-+sqrt%28x%29+=+6

To solve, isolate the radical and square both sides:

x-6+=+sqrt%28x%29
x%5E2-12x%2B36+=+x
x%5E2-13x%2B36+=+0
%28x-4%29%28x-9%29+=+0

The possible solutions are x=4 and x=9.

But since, in solving the equation we squared both sides of the equation, we need to check for extraneous solutions.

And indeed x=4 does not work: 4+-+sqrt%284%29+=+4-2+=+2 is not equal to 6.

But x=9 works: 9+-+sqrt%289%29+=+9-3+=+6.

So the single solution is x=9.

Answer by ikleyn(52784) About Me  (Show Source):
You can put this solution on YOUR website!
.
the difference between an integer and its square root is 6. find the integer.
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Since the difference between an integer and its square root is 6, the square root itself is an integer.


Let x be the square root; then the integer under the question is x^2, and the condition says


x^2 - x = 6,   or  x*(x-1) = 6.


Hence, x is the divisor of 6, such that (x-1) is the divisor of 6 too; and the product of these two consecutive integers is 6.


There is only one such a divisor of 6: it is 3.  


Hence, the integer under the question is 9.

Solved.