solve (2-x²) (1+x²+x³) + (4+x) (3-x-x²)
please give right answer
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solve (2-x²) (1+x²+x³) + (4+x) (3-x-x²)
please give right answer
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To answer for your question, first you need to divide both sides by x^2. This will allow you to get rid of x^2.
So here is what you are expected to get.
x( x-1) ( x+1 ) = ( x- 1 )^2
Then you would divide both sides by ( x-1 ) to make life easier, and notice that ( x- 1 )^2 is the same as multiplying (x- 1 ) twice. Like this : (x- 1 ) ( x- 1 )
This allow you to cancel out ( x- 1 ) from both sides.
And now the question get simpler and become easier to solve. This is what you should get.
x( x+1) = ( x-1)
The next step would be to Expand the bracket out .
So you have this equation, hopefully.
x^2 + x = x - 1
Bring all the term into one side and makes the equation equal to 0.
You equation will then becomes like this.
x^2 + x - x + 1 = 0 , and the two x(s)will then cancel out and you would be left with an equation :
x^2 + 1= 0
Subtract one from both sides in order to be left with x^2 on one side.
x^2 + 1 - 1 = 0 - 1
x^2 = -1
So now you have x^ 2 on the one side and -1 on the other side.
The final step to get x alone, to square root both sides, to get x by itself.
Square Root of (x )^2 = plus or minus square Root of (-1)
In this case you can’t apply square root on negative numbers. This isn’t possible( mathematically speaking).
E.g if you square any number whether positive or negative, the answer will always be positive so, there is no way that you could square a number and get negative number as a result( in real numbers ).
But COMPLEX Numbers does allow you to have negative number under root sign.
Then if you allow complex number to involve your equation, you will have the answer for x.
X = plus or minus square root of -1. Mathematicians respresent root -1 as lowercase i ( short for imagine numbers )
THEREFORE : The values for x would be plus or minus root -1 or just plus or minus i ( if you want to be more specific )
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