x 2 − 3 x + 2 x 3 − 5 x 2 + 8 x − 4 IF THE VALUE OF ABOVE EQUATION IS 1 THEN,
FIND ALL THE VALUES OF X
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Even I did the same ^_^
We can factor the given equation as ( x − 1 ) ( x − 2 ) ( x − 1 ) ( x − 2 ) 2 = 1 .
Now for both x = 1 and x = 2 both the numerator and denominator equal 0 , hence making the expression on the LHS indeterminate. Thus there are points of discontinuity for this expression at x = 1 , 2 .
For x = 1 , 2 we can cancel terms to obtain the simplified equation x − 2 = 1 ⟹ x = 3 .
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There was no need to solve, just observe that all the other 4 options contained 1,2 or both which isn't possible as the denominator then would be 0.