the least value of n for which n ^4 - 109 n ^2 + 900 is negative

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Akira23

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the least value of n for which n ^4 - 109 n ^2 + 900 is negative.
this isn’t my homework question, I just wanted to know how to solve it
I felt it should be like n ^4 > 109 n ^2 + 900
But I don’t know how to go about it
 
I would factor the expression as follows:

[MATH](n^2-100)(n^2-9)[/MATH]
And then further factor as:

[MATH](n+10)(n-10)(n+3)(n-3)[/MATH]
Now you have 4 roots which divide the real number line into 5 intervals...and since none of the roots are repeated, the sign of the expression will alternate across all intervals. Can you give the intervals where the expression is negative?
 
Akira23 said:
The answer is 12 , I don’t get it from the explanation. I know the answer but not the solution , please help if you know how to go about the answer
Click to expand...

The expression is positive for \(n=12\).

Can you give the intervals on which the expression is negative?
 
I saw that you briefly posted a claimed answer of 12, then it went away. That is clearly not the answer, since when n=12, n^4 - 109 n^2 + 900 = 5940.

But this raises another question. Since you didn't post the entire question, please do so; perhaps n is assumed not just to be an integer, but a positive integer.
 
Akira23 said:
the least value of n for which n ^4 - 109 n ^2 + 900 is negative.
I felt it should be like n ^4 > 109 n ^2 + 900
Click to expand...
In your expression it seems that you were thinking that the expression was greater than 0, but when you moved two of the terms to the other side they went from having different signs to having the same signs. This is a sign that something went wrong!
 
Akira23 said:
the least value of n for which n ^4 - 109 n ^2 + 900 is negative.
this isn’t my homework question, I just wanted to know how to solve it
I felt it should be like n ^4 > 109 n ^2 + 900
But I don’t know how to go about it
Click to expand...
Well, no, it would be n^4> 109 n^2- 900!

But I wouldn't do that. Instead I would let m= n^2 so that n^4- 109n^2+ 900 becomes the quadratic m^2- 109m+ 900. That factors (thanks to MarkFL for doing the hard work for me) as (m- 100)(m- 9) and is 0 (so can change from negative to positive and vice-versa) where m= n^2= 100 and m= n^2= 9 so where n= 10, -10, 3, and -3. n^4- 109n^2+ 900= (n+ 10)(n+ 3)(n- 3)(n- 10). If n< -10, all four factors are "n- a larger number" so are all negative. The product of 4 (an even number) of negatives is positive. But if -10< n< -3, n+ 10 is positive while the other three factors are still negative. The product of three negatives is negative. Similarly, for -3< n< 3, two factors, (n+ 10) and (n+3), are positive while the other two, (n-3) and (n- 10), are positive. That product is positive. For 3< n< 10, there are three positive factors and one negative. That product is negative. Finally, for n> 10, all four factors are positive so the product is positive.

The smallest integer value for which this is negative is the smallest integer larger than -10 which is -9.
 
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