It's zeroes
Let p and q be the zeroes of the polynomial f ( x ) = x 2 − 5 x + 4 . Find ∣ ∣ ∣ ∣ p 1 + q 1 − 2 p q ∣ ∣ ∣ ∣ .
Notation: ∣ ⋅ ∣ denotes the absolute value function
The answer is 6.75.
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3 solutions
x 2 − 5 x + 4 = 0
Factor.
( x − 1 ) ( x − 4 ) = 0
x − 1 = 0 or x − 4 = 0
x = 1 or x = 4
So, ∣ ∣ ∣ ∣ p 1 + q 1 − 2 p q ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ 1 1 + 4 1 − 2 ( 1 ) ( 4 ) ∣ ∣ ∣ ∣ = ∣ − 6 . 7 5 ∣ = 6 . 7 5
p = 1 , q = 4 , ∣ p 1 + q 1 − 2 p q ∣ = ∣ 1 + 4 1 − 8 ∣ = 6 . 7 5 .
It was not a clever way, you can do it better with out finding the zeroes and using vieta's formula instead
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For this problem, it suffices to use the roots directly.
As it has the obvious roots, you should use substitution rather then use vieta's formula
You should set a problem without real roots or use a higher degree polynomial, then the use of Vieta's formula will be much convenient.
We can solve this problem without solving for x .
By Vieta's formula , we have { p + q = 5 p q = 4 . And ∣ ∣ ∣ ∣ p 1 + q 1 − 2 p q ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ p q p + q − 2 p q ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ 4 5 − 2 ( 4 ) ∣ ∣ ∣ ∣ = ∣ − 6 . 7 5 ∣ = 6 . 7 5 .