Question

According to the Rational Root Theorem, the following are potential roots of f(x) = 60x2 – 57x – 18. -6/5, -1/4, 3, 6 Which is an actual root of f(x)?

Answer 1

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. 

Answer 2

The actual root of the function $60x^2 -57x -18$ is $\dfrac{-1}{4}$.

What is the rational root theorem?

According to the rational root theorem, each rational solution $x=\dfrac{p}{q}$ 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,

 $f(x) = 60(\dfrac{-6}{5})^2 - 57(\dfrac{-6}{5}) - 18 = 136.8$

This is not the actual root.

B.)  -1/4,

$f(x) = 60(\dfrac{-1}{4})^2 - 57(\dfrac{-1}{4}) - 18 = 0$

This is the actual root.

C.) 3

$f(x) = 60(3)^2 - 57(3)- 18 = 351$

This is not the actual root.

D.) 6

$f(x) = 60(6)^2 - 57(6)- 18 = 1800$

This is not the actual root.

Hence, the actual root of the function $60x^2 -57x -18$ is $\dfrac{-1}{4}$.

Learn more about the Rational root theorem:

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