Are you sure they exist?
If the square of x and the square root of x are equal to each other, but not equal to x , then what is the value of x 2 + x ?
The answer is -1.
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4 solutions
Nice simple solution. voted up. I have also given a different approach with a little higher math.
What am I missing here. With x=-1 doesn't the first equation imply 1=i?
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x is a complex number. It's x^2 + x that equals -1.
well explain
Just wondering.. If x^2 + x + 1 = 0 then x = w = e^i2pi/3 & x = w^2 = e^i(-2pi/3) Both of which don't satisfy the original equation Since for w: w^2 = e^i(4pi/3) = e^i(-2pi/3) & root(w) = e^i(pi/3) thus, w^2 is not equal to root(w) Similarly for the other root..
X 2 = X 2 1 , ⟹ X 2 3 = 1 , ∴ X = 1 3 2 = ω 2 . ∴ X 2 + X = ( ω 2 ) 2 + ω 2 = ω + ω 2 = − 1 ω is cube root of 1.
Yup, I too did the same way
x 2 = x
x 3 = 1
the roots:
x 1 = e 3 2 π i x 2 = e − 3 2 π i x 3 = 1 , which we rule it out
therefore x 2 + x = x 1 + x 2 = − 1
x^3 = 1 cant be the same as x^2 = sqrt(x) as it remove the possibility of x=0
x 2 = x , so squaring both sides, we get that x 4 = x
Since x = 0 , x 3 = 1
Now we wish to find x 2 + x = x 2 + x + 1 − 1 = x − 1 x 3 − 1 − 1
But since x 3 = 1 , this is simply − 1