Polynomial function
x 7 + 1 4 x 5 + 1 6 x 3 + 3 0 x − 5 6 0 .
How many real solutions does the equation have?
(Its an AIEEE 2008 problem)
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3 solutions
We can simply proceed like ths : f ′ ( x ) = 7 x 6 + 7 0 x 4 + 4 8 x 2 + 3 0 > 0 Therefore function is always increasing...
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Very good approach kinshuk
But remember to show that there is at least one real root.
that is an excellent approach :D
Aditya. Its not from pace booklet. Its from AIEEE Exam
I'm sorry to ask, but what's this condition? I haven't seen this thing ever. :(
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It's Descarte's rule of sign changes , which states that if there are n sign changes in f ( x ) (the + sign changing to - and - sign changing to +), then there will be n or n − 2 or n − 4 or n − 6 or .... n ( m o d 2 ) number of positive roots.
Then replace x with − x , then in f ( − x ) , if you get k sign changes, then there will be k or k − 2 or k − 4 or .... or k ( m o d 2 ) number of negative roots.
Also, check whether 0 is a root or not.
This condition can give information about real roots, because if at all there is any real root then either it will be -ve or +ve or 0.
It's from AIEEE-2008.The pace booklet too mentions it.
Let f ( x ) = x 7 + 1 4 x 5 + 1 6 x 3 + 3 0 x − 5 6 0 f ( 0 ) = − 5 6 0 Simply, as the powers are odd, if we put a x=negative number , we will get negative number ,i.e. there is no change in sign of f ( x ) .
f ( 1 ) is also less than 0
but f ( 2 ) is positive.
And putting x greater than 2, we'll get Positive f ( x )
That means sign of f ( x ) changes only between x=1 and x=2.
So, f ( x ) = 0 for only 1 value of x
Use the Descarte's condition of real (positive and negative roots)
We know that if there are 2 n sign changes in f ( x ) starting from the first term, then it may have 2 n or 2 n − 2 or 2 n − 4 so on upto 0 number of POSITIVE roots.And if there are 2 n + 1 sign changes then there could be 2 n + 1 or 2 n − 1 or 2 n − 3 so on up to 1 positive roots. And applying the same thing to f ( − x ) gives the number of negative roots.
Observe that here, starting from the first term, + changes to - and - changes to + exactly ONCE.
And the number of changes of sign in f ( − x ) here is 0.
This implies that the equation has 1 positive root and 0 negative roots. Also, 0 is not a solution obviously.
Thus this equation has 1 real root and only 1 real root.