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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)?

-6/5
-1/4
3
6

Respuesta :

daniel42steelers
The correct answer for the question that is being presented above is this one: "-4 and 3." According to the Rational Root Theorem, the following are potential f(x) 2x2 +2x 24. The actual roots of the quadratic equation based on the Rational Root Theorem are -4 and 3.

We have been given the function [tex]f(x) = 60x^2 - 57x -18[/tex]

Now, the given potential roots are -6/5, -1/4, 3, 6. In order to find the actual root of this function, we substitute the root in the given function and if we get zero, then that would be the actual zero.

[tex]f(-6/5)=60(-6/5)^2-55\cdot\frac{-6}{5} -18= \frac{684}{5}\neq 0\\\\f(-1/4)=60(-1/4)^2-55\cdot\frac{-1}{4} -18=0\\\\\\f(3)=60(3)^2-57(3)-18=351\neq 0\\\\f(6)=60(6)^2-57(6)-18=1800\neq 0[/tex]

We got zero for the value -1/4.

Hence, the actual root of f(x) is -1/4.

B is the correct option.

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