Algebra Question #13: Inspired by Mei Li
Find the number of integer x satisfying ( x 2 − 5 x + 6 ) x 2 − 6 x + 8 = 1
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4 solutions
zero to the power zero is not defined then why does the calculator show it as 1
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lim x -> 0 x^x = 1
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0^0 = 0^(a - a) = 0^a/0^a = indeterminate and not 1.
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thanks for clearing my concept !!
− 9 ∗ − 9 = 3 i ∗ 3 i = − 9 , but − 9 ∗ − 9 = 8 1 = 9 = − 9 , so some rules have exceptions.
Since when are calculators perfect - there are plenty of examples on this site where calculators should not be trusted. Most Calculators use short cuts - which don't work in all cases : 0^0 is unsually considered to be undefined : x a = 0 : x = 0 a x = 1 : x = 0 so x x = ? : x = 0 which rule does one use ?
Perhaps it should be 1, and is in come contexts, but it's not; and calculators make mistakes.
Calculators are not mathematicians.
Because some calculators are wrong - undefined means it could be any value - depending how you get there - lim x -> 0^ x = 0
but we also know that x^0 = 1 for non zero x - so is 0^0 = 1 or 0; there are also definitions which can make it 3 or 7 or 1 million 350 thousand or any value you want.
0^0 is not define
Did the same.
The only possibilities are if the exponent = 0, or the argument is +/- 1. (if -1, the exponent must be even). Equating the exponent to 0 gives x =2, x= 4, However, 2 must be eliminated since the argument = 0.Equating the argument to |1| produces no integer solutions, so there is only 1. Ed Gray
The left hand side will be equal to 1 when the exponent is zero:
x 2 − 6 x + 8 = 0
This has x = 2 and x = 4 as solutions.
Of course, for the left hand side to be defined, x 2 − 5 x + 6 should NOT be zero (otherwise we would end up with 0 0 , which is undefined), so:
x 2 − 5 x + 6 = 0
This solves to x = 2 and x = 3
Hence our original solution cannot include the number 2, and we are left with the number 4 as a solution.
So there is just o n e possible value for x .
( x 2 − 5 x + 6 ) x 2 − 6 x + 8 = 1
This will be correct only in 2 cases
1- If x 2 − 5 x + 6 = 1 in this case the results aren't integers
2- If x 2 − 6 x + 8 = 0 which will result in x = 4 a n d x = 2
In case of x = 2 the equation will be 0 0 which is an indefinable quantity
So only if x = 4 that it'll be 2 0 = 1
( x 2 − 5 x + 6 ) x 2 − 6 x + 8 = 1 if and only if x 2 − 6 x + 8 = 0 = > x = 2 o r x = 4 x 2 − 5 x + 6 = 1 = > x i s n o t a n i n t e g e r . For even exponent x 2 − 5 x + 6 = − 1 = > x i s n o t a n i n t e g e r . But, using substitution, the only answer is x = 4, because if x = 2 the original equation will become 0 0 which is not equal to 1