Respuesta :
As for Rational Root Theorem, we substitute the given choices in the equation and if the value of f(x) is equal to 0 then, the certain choice is a root of the equation.
(1) -6/5 : f(x) = 60(-6/5)^2 - 57(-6/5) - 18 = 136.8
(2) -1/4: f(x) = 60(-1/4)^2 - 57(-1/4) - 18 = 0
(3) 3: f(x) = 60(3)^2 - 57(3) - 18 = 351
(4) 6: f(x) = 60(6)^2 -57(6) - 18 = 1800
Thus, the answer is the second choice.
(1) -6/5 : f(x) = 60(-6/5)^2 - 57(-6/5) - 18 = 136.8
(2) -1/4: f(x) = 60(-1/4)^2 - 57(-1/4) - 18 = 0
(3) 3: f(x) = 60(3)^2 - 57(3) - 18 = 351
(4) 6: f(x) = 60(6)^2 -57(6) - 18 = 1800
Thus, the answer is the second choice.
The actual root of the function [tex]60x^2 -57x -18[/tex] is [tex]\dfrac{-1}{4}[/tex].
What is the rational root theorem?
According to the rational root theorem, each rational solution [tex]x=\dfrac{p}{q}[/tex] is written in the lowest terms so that p and q are relatively prime.
Which is an actual root?
To check which is an actual root of the given function, substitute the value of x in f(x),
A.) -6/5,
[tex]f(x) = 60(\dfrac{-6}{5})^2 - 57(\dfrac{-6}{5}) - 18 = 136.8[/tex]
This is not the actual root.
B.) -1/4,
[tex]f(x) = 60(\dfrac{-1}{4})^2 - 57(\dfrac{-1}{4}) - 18 = 0[/tex]
This is the actual root.
C.) 3
[tex]f(x) = 60(3)^2 - 57(3)- 18 = 351[/tex]
This is not the actual root.
D.) 6
[tex]f(x) = 60(6)^2 - 57(6)- 18 = 1800[/tex]
This is not the actual root.
Hence, the actual root of the function [tex]60x^2 -57x -18[/tex] is [tex]\dfrac{-1}{4}[/tex].
Learn more about the Rational root theorem:
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