Vieta's Derivative II -- My 500-follower problem
The function
x
5
+
6
x
4
−
1
8
x
3
−
1
0
x
2
+
4
5
x
−
2
4
has only four distinct real roots:
α
,
β
,
γ
and
δ
(in no particular order). If
f
′
(
x
)
is the first derivative of
f
(
x
)
. Evaluate
f ′ ( α ) + f ′ ( β ) + f ′ ( γ ) + f ′ ( δ )
Inspiration: This is exactly the problem titled Inspired by Vieta's Derivatives with a twist.
The answer is 4797.000.
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2 solutions
Thanks for the answer. It helps if you plot a graph. We can do it quite fast using a spreadsheet.
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My favorite part of the problem is the clause "with a twist": it is quite literally a twist, as the × sign has been rotated 4 5 ∘ .
Thanks! I only use them as a last resort or if I'm feeling lazy.
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I understand. You like to exercise your brain first. But for me I make quite a bit or careless mistakes, therefore, need to use a surer way.
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@Chew-Seong Cheong – To each his own! Big fan of your answers!
is there any easy way to do it ....?
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I doubt there is one. One slightly simpler way to factor out ( x 2 − 3 ) after ( x − 1 ) 2 has been factored out is to consider the equation x 3 − 3 x = 8 x 2 − 2 4 ⇒ x ( x 2 − 3 ) = − 8 ( x 2 − 3 ) . But this only simplify the overall working by a little.
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i did the same and got wrong due to calculation mistakes!!
these type of question gets us tired and tedious ....!!
but still a good question indeed!
good question but far too tedious.............
This is in response to Aman Gautam 's comments. I hope this is helpful.
I find it easy to solved it using an Excel spreadsheet. Which I am going to show here.
First we can plot the graph of f ( x ) = x 5 + 6 x 4 − 1 8 x 3 − 1 0 x 2 + 4 5 x − 2 4 . From the calculated data, we see that there are four real roots − 8 , ( − 1 . 7 5 , − 1 . 5 ) , 1 , ( 1 . 7 5 , 1 . 5 ) .
By changing the starting
x
value (Cell A1) and the graph scaling (Cell C1), we can zoom in around
x
=
1
we see that
x
=
1
is the double root (see below). We can confirm this by putting values in
f
′
(
x
)
=
5
x
4
+
2
4
x
3
−
5
4
x
2
−
2
0
x
+
4
5
. (Notice the formula in Cell B1.)
Then we can use spreadsheet to do long division by the known factors x − 1 , x − 1 , x − 8 , as follows:
Row 1 is the coefficients of f ( x ) . Column A is the factors. Row 3 is the quotient of x − 1 f ( x ) ⇒ f ( x ) = ( x − 1 ) ( x 4 + 7 x 3 − 1 1 x 2 − 2 1 x + 2 4 ) .
Similarly, Row 5: ⇒ f ( x ) = ( x − 1 ) 2 ( x 3 + 8 x 2 − 3 x − 2 4 )
And, Row 7: ⇒ f ( x ) = ( x − 1 ) 2 ( x + 8 ) ( x 2 − 3 )
Therefore, the remaining two roots are ± 3
Of course, we can use the speadsheet to calculate the value of f ′ ( α ) + f ′ ( β ) + f ′ ( γ ) + f ′ ( δ ) .
just Brilliant!
By Rational Root Theorem and the suggestion of the description, by perform long division twice, we can see that x = 1 is double root of the equation. Therefore leaving the cubic polynomial ( x 3 + 8 x 2 − 3 x − 2 4 ) .
A few tedious trial and error for factors of ± 2 4 gives x = − 8 as another root for the cubic polynomial. So the equation factors completely to ( x − 1 ) 2 ( x + 8 ) ( x 2 − 3 ) . Thus the value of the expression in question is simply f ′ ( 1 ) + f ′ ( − 8 ) + f ′ ( 3 ) + f ′ ( − 3 ) .
With f ′ ( x ) = 5 x 4 + 2 4 x 3 − 5 4 x 2 − 2 0 x + 4 5 , we can easily obtain f ′ ( 1 ) = 0 and with enough patience, we get f ′ ( − 8 ) = 4 9 4 1
And f ′ ( x ) + f ′ ( − x ) = 2 ( 5 x 4 − 5 4 x 2 + 4 5 ) , with x = 3 , we have f ′ ( 3 ) + f ′ ( − 3 ) = 2 ( 5 ⋅ 3 2 − 5 4 ⋅ 3 + 4 5 ) = − 1 4 4
The answer is 0 + 4 9 4 1 − 1 4 4 = 4 7 9 7