Vieta's was fine, until this problem...
The equation
x 6 − 7 x 5 + 8 x 4 − 9 x 3 + 1 0 x 2 − 1 1 x + 1 2 = 0
has 6 (possibly complex) roots a 1 , a 2 , . . . , a 6 . Determine the value of
∣ ∣ ∣ ∣ ∣ i = 1 ∏ 6 ( a i − 1 ) ∣ ∣ ∣ ∣ ∣
The answer is 4.
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3 solutions
FYI, re Latex, it's generally better to use \left and \right to let Latex figure out the size of the brackets automatically, instead of setting it manually (unless you really need it to stress a certain part of the equation).
Also, using \ [ \ ] puts your equation in \displaystyle already, so that part of the code is unnecesssary. \displaystyle is used for
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Oh, I thought \ [ \ ] put my equation in the middle and \displaystyle is for the i = 1 be under the ∏ and 6 on it. Normally I don't care how the code is, I only want to make sure my problem is well-presented so I end up making these unnecessary codes but I will be more alert next time. Thanks for your LaTeX guidance! :)
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No worries. The minute technicalities of Latex takes a while to understand and get used to. You are doing great!!
When I first started out, I also used whatever methods I could to get it to display the way I wanted, which included forcing spaces in my equations, combining symbols in weird ways, etc.
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@Calvin Lin – @Calvin Lin Can you tell me where I could learn all LaTeX commands because I face a lot of problems while uploading a question and am not able to upload complicated ones!
Gosh dang it! I just bashed it out... :(
Another method could be to substitute : [ x by (t+1) ] in the equation above, then the roots of that equation would be: a 1 − 1 , a 2 − 1 , a 3 − 1 , a 4 − 1 , a 5 − 1 , a 6 − 1 Whereupon the constant term of our new formed equation would be our answer.....
But darn it!!! Your solution is much cooler than mine.... :)
Nice problem! Did the same way :)
Oh no! I forgot the 1 and took it all wrong! Learning from mistakes!
but( 1 − a 1 )( 1 − a 2 ).....( 1 − a 6 ) =4. Then how ( a 1 − 1 )( a 2 − 1 ).... ( a 6 − 1 )=4
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because -1(1-a1) = a1-1 , and you do that to an even number of numbers so it cancels out into 1
Very simple problem , I made it incorrect
Let b i = a i − 1 . Thus we are now finding the absolute value of ∏ i = 1 6 ( b i ) .
We will rewrite the equation as ( b + 1 ) 6 − 7 ( b + 1 ) 5 + 8 ( b + 1 ) 4 − 9 ( b + 1 ) 3 + 1 0 ( b + 1 ) 2 − 1 1 ( b + 1 ) + 1 2 = 0
According to the Vieta's Theorem, what we care about is ∏ i = 1 6 ( b i ) and that is equal to the constant in the equation.
So, ignoring the variables, the constant is 1 − 7 + 8 − 9 + 1 0 − 1 1 + 1 2 = 4 . Also note that we don't have to worry about the signs(+,-) of 4 because the question is asking for the absolute value of it.
Sorry for my poor latex.
This is how I did but I admit the non-Vieta solution is pretty elegant too.
The numbers a i − 1 are the roots of f ( x + 1 ) . And the product of all roots of a polynomial equals the constant term which in this case can easily be calculated by summing each coefficient. So the answer is 4
Let
f ( x ) = x 6 − 7 x 5 + 8 x 4 − 9 x 3 + 1 0 x 2 − 1 1 x + 1 2 = ( x − a 1 ) ( x − a 2 ) . . . ( x − a 6 )
If we let x = 1 , we have
f ( 1 ) = 1 − 7 + 8 − 9 + 1 0 − 1 1 + 1 2 = ( 1 − a i ) ( 1 − a 2 ) . . . ( 1 − a 6 )
Hence,
( a 1 − 1 ) ( a 2 − 1 ) . . . ( a 6 − 1 ) = i = 1 ∏ 6 ( a i − 1 ) = 4 .