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5 solutions
Lol. I should have had faith in my intuition.
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This question was intentionally tricky. Click through on the inspiration, which deals with a similar scenario, though not in such an obvious manner.
These arguments are false. 0/0 is permitted as an indeterminate form. One must look to the limiting case as x tends to 0. In this case l'Hopital's rule applies and one may write lim {x\to 0}(x/x)=lim {x\to 0}(1/1)=1. Sorry guys but you're wrong. There is one solution and its zero (twice, degenerately).
we don't divide i by itself to get 1?
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Note that i 2 + i i = − 1 + 1 = 0 = 1 .
So yes, when we divide i by itself we get 1. However, i is not a solution to the equation.
I didn't notice COMPLEX SOLUTION.
I don't follow the logic here. Why is it that because x = 0, there are no complex solutions? I figured because x cannot be zero and it's quadratic, that there'd still need to be 2 solutions, and since 0 can't be one of those, it led me to believe that there must be some imaginary roots?
Although, if you calculate the discriminant, you end up with 0, which implies a multiplicity of 2 for x = 0, but that still can't be the solution... any insight appreciated.
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Note that the question asks for "how many complex solutions" , and not "what are the complex solutions". This is why the answer is "none", instead of "0".
It cannot be 0 because you cannot divide by 0. In particular, 0 0 is undefined, and is not equal to 1.
If you went to multiply throughout by x , you will obtain the equation x 2 = 0 . It has a repeated root (like you mentioned) of 0, which is not a valid answer.
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As the question asked "how many complex solutions" the function should be f(z)=z^2+(z/z) which has a removable singularity at z=0 , when removed gives f(0)=1 so actually there is a solution :-s
what if x is infinitely small no!! then its correct. So it should be specified that only integers, else the question is wrong
that's interesting. I guess we don't divide i by itself to get i because i/i = sqrt(-1/-1)=sqr(-1) which is not real, and 1 is real
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minus divided by a minus is a positive ?
x 2 + x x = 1
We can see that the only restriction is that x = 0 , because then we'd divide by 0, which is undefined. If we assume this, we can do the following steps without any problems.
x 2 + x x = 1
⟺ x ( x + x 1 ) = 1
⟺ x + x 1 = x 1
⟺ x = 0
From the last statement we get the information that the equation x 2 + x x = 1 can only be true, if and only if x = 0 . This is a contradiction to our restriction for x = 0 . Thus, there are no complex solutions to this equation.
since x/x = 1
x²+1=1
x²=0
x=0 but x can't be 0......bcoz if x=0 x/x would become undefined.
Therefore, no solution.
x/x = 1 So we get: x^2 + 1 = 1 x^2 =1-1 x^2 = 0 Since x can never be 0. There would be no Solution
x 2 + x x = 1 x x 3 + x = 1 x 3 + x = x x 3 = 0 x = 0 . or, x 2 + x x = 1 x 2 + 1 = 1 x 2 = 0 x = 0 So, it has only one real solution ‘0’ and no complex solution.
I was thinking x= +/- ((sqrt 2) - 1)
x cannot be 0 , otherwise the LHS will be undefined.
Since x = 0 , x x = 1 . The equation becomes x 2 + 1 = 1
⟹ x 2 = 0
⟹ x = 0 . However x cannot be 0 , therefore there are no complex solutions.