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(1) gives us x=0,-3,2EbrahimHashem wrote:What is the value of x ?
(1) x^3 = 6x - x^2
(2) 14x = x^3 + 5x^2
HAVE FUN!
Your answer is correct! but what's wrong with the first answer provided by"kyabe"? is there anything we both are missing?qwe12 wrote:(1) gives us x=0,-3,2EbrahimHashem wrote:What is the value of x ?
(1) x^3 = 6x - x^2
(2) 14x = x^3 + 5x^2
HAVE FUN!
(2) gives us x=0,2,7
from (1) and (2) we get x=0,2
since the question asks what *is* (singular) the value of x, the answer must be (E), since from neither (1) or (2) can we get one single value for x.
(E)
Ebrahim the answer posted by qwe12 should be correct as when I used deduced the solution I cancelled x from LHS and RHS. Now when we cancel X, 0 can be one of the factors. And as said by qwe12 after combining i and ii we are left with two values of x ie 0 and 2. Hence its E. Does that makes sense?EbrahimHashem wrote:Your answer is correct! but what's wrong with the first answer provided by"kyabe"? is there anything we both are missing?qwe12 wrote:(1) gives us x=0,-3,2EbrahimHashem wrote:What is the value of x ?
(1) x^3 = 6x - x^2
(2) 14x = x^3 + 5x^2
HAVE FUN!
(2) gives us x=0,2,7
from (1) and (2) we get x=0,2
since the question asks what *is* (singular) the value of x, the answer must be (E), since from neither (1) or (2) can we get one single value for x.
(E)
