Not all quartics are as difficult to solve...
Find the mean of all real solutions to: n 3 − 4 = n 4 − 5 n 2 + 4 n
The answer is 0.25.
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2 solutions
You could have just used Vieta's formulas.
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No. Since he is asking for the mean of real solutions, you do not know what the sum of the imaginary solutions are. It just happened that the (corrected) polynomial has no imaginary solutions.
Could you explain further please?
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Vieta's formulae are the generalisation of my solution. It basically states that the coefficients are elementary symmetric functions of the roots. However one does not need the formulae to solve this problem as it is simpler to see the permutation yourself rather than blackboxing the solution.
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@A Former Brilliant Member – You should also note that elementary symmetric functions are an important concept in commutative algebra and their properties are generally difficult to prove. However this is a lot of abstract nonsense and really will not help you solving polynomials. The one thing you should take from this is the permutation of the roots.
I accidentally posted my solution as private so I shall post here instead:
All monic quartics with roots a , b , c and d can be written as ( x − a ) ( x − b ) ( x − c ) ( x − d ) = x 4 − ( a + b + c + d ) x 3 + ( a b + a c + a d + b c + b d + c d ) x 2 + ( a b c + b c d + c d a + d a b ) x + a b c d Now the expression we would like is 4 a + b + c + d notice that this is just the unsigned second coefficient! So after rewriting the quartic equal to zero we have our answer as 4 1 = 0 . 2 5 .
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Yeah but you need to make sure that all roots are real. But they're actually all real. XD
Try synthetic division and you'll see the famous quadratic equation x 2 − x − 1 = 0 . And we're done.
Thanks for the excellent solution Ali! It definitely shows how different methods can be a lot more elegant :)
This problem is wrong, see that the person who has written the solution (the person who made the problem) is having a little misconception.
( n 2 − 4 ) ( n 2 − 1 ) is actually n 4 − 5 n 2 + 4 and from the question, it's seen to be n 4 − 5 n 2 − 4 . Either you need to change the answer or change the sign of 4 .
Please look into this !
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Thanks for pointing out the error.
I have updated the question, since otherwise the equation is not easily solved.
rearrange the following into a general equation
a{ x }^{ 4 }\quad +\quad b{ x }^{ 3 }\quad +\quad c{ x }^{ 2 }\quad +\quad d{ x }^{ 1 }\quad +\quad e
Sum of roots is -b/a
in our case a=1 b=-1
and the degree of equation is 4. So number of roots will be 4
Means = -(1/-1)/4
We need to show that all roots are real
To solve a quartic, we need to use some intuition; let's try to get something we can factorise. Move the +4 to the RHS and the +4n to the LHS, then we get n 3 − 4 n = n 4 − 5 n 2 − 4 n ( n 2 − 4 ) = ( n 2 − 4 ) ( n 2 − 1 ) ( n 2 − 4 ) ( n 2 − n − 1 ) = 0 Hence the solutions are 2, -2, 2 1 + 5 , 2 1 − 5 , and the mean is 0.25.