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# solve the following equation: x^2-6x+[x]+7=0. how to solve this?

Arun
25763 Points
3 years ago
Dear student

We can always write
[X] = X - r
where 0
Then
X^2 - 6X + [X] + 7 = 0
becomes
X^2 - 5X + 7 - r = 0
or, by completing the square:
(X-5/2)^2 + 3/4 - r = 0
If
|X - 5/2| >= 1/2,
then we would have
(X-5/2)^2 >= 1/4
and therefore
r = (X-5/2)^2 + 3/4 >= 1,
which is a contradiction. Therefore, we must have
|X - 5/2|
or  2
Now it's easier. This means [X] = 2, so
X^2 - 6X + 2 + 7 = 0
(X-3)^2 = 0
X = 3.
HOWEVER, X=3 contradicts [X]=2, so in the end, there is no solution.

Regards
suvidhi mehta
292 Points
3 years ago
you can also make th graph and check you will get no solutions
[x] = 6x – x^2 – 7
hope it helps you..........
Soumendu Majumdar
159 Points
2 years ago
Dear Student,
$x^2 - 6x + [x] + 7 = 0$
$\Rightarrow (x-2.5)^2 + 0.75 - (x-[x]) = 0$
so {x} $\geq 0.75$
But for {x} = 0.75 the value of x = 2.5 , which is impossible
So there is no value of x for which this equation is satisfied.

Hope it helps!