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26 Apr 2012, 23:45
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Given that (x^3+4x^2+4x)/x = 9. What is x?
I solved this as
x(X^2+4x+4)/x = 9
(x+2)(x+2) = 9
so x = 9/2
But the actual answer is x={-5,1}
while achieving the above answer, they have not canceled x like i have done. But approached it as X^3+4x^2+4x = 9x and have solved.
What is wrong with my approach. I am looking for any technical reasons as to why it should not be approached the way i have.
x(X^2+4x+4)/x = 9
(x+2)(x+2) = 9
so x = 9/2
But the actual answer is x={-5,1}
while achieving the above answer, they have not canceled x like i have done. But approached it as X^3+4x^2+4x = 9x and have solved.
What is wrong with my approach. I am looking for any technical reasons as to why it should not be approached the way i have.
27 Apr 2012, 00:17
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mohankumarbd
Given that (x^3+4x^2+4x)/x = 9. What is x?
I solved this as
x(X^2+4x+4)/x = 9
(x+2)(x+2) = 9
so x = 9/2
But the actual answer is x={-5,1}
while achieving the above answer, they have not canceled x like i have done. But approached it as X^3+4x^2+4x = 9x and have solved.
What is wrong with my approach. I am looking for any technical reasons as to why it should not be approached the way i have.
I solved this as
x(X^2+4x+4)/x = 9
(x+2)(x+2) = 9
so x = 9/2
But the actual answer is x={-5,1}
while achieving the above answer, they have not canceled x like i have done. But approached it as X^3+4x^2+4x = 9x and have solved.
What is wrong with my approach. I am looking for any technical reasons as to why it should not be approached the way i have.
Given that (x^3+4x^2+4x)/x = 9. What is x?
Since \(x\) is in denominator then we know that it cannot equal to zero, so we can reduce it the way you did: \(\frac{x^3+4x^2+4x}{x}=9\) --> \(x^2+4x+4=9\) --> \(x^2+4x-5=0\) --> \((x+5)(x-1)=0\) --> \(x=-5\) or \(x=1\).
Now, the way you are doing is wrong since if you substitute \(x=\frac{9}{2}\) in \((x+2)^2=9\) equation just won't hold true. Though you can proceed this way too: \((x+2)^2=9\) --> \(|x+2|=3\) --> \(x=-5\) or \(x=1\).
Hope it helps.
29 Apr 2012, 03:10
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(x^3+4x^2+4x)/x = 9. What is x?
it can be written as:
x(X^2+4x+4)/x = 9
or,
x^3+4x^2+4x=9x
or
x(X^2+4x-5)=0
x(x-1)(x+5)=0
so
possible solution is x =0,1 & -5
but if we put x = 0 in original equation , it will not be defined.
hence,
Answer is 1 & -5 .
it can be written as:
x(X^2+4x+4)/x = 9
or,
x^3+4x^2+4x=9x
or
x(X^2+4x-5)=0
x(x-1)(x+5)=0
so
possible solution is x =0,1 & -5
but if we put x = 0 in original equation , it will not be defined.
hence,
Answer is 1 & -5 .
