Question

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

Answer 1

Answer: The correct option is third, i.e., $-\frac{1}{4}$.

Explanation:

The given function is,

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

The Rational Root Theorem states that the rational roots are in the form of,

$r=\frac{\text{factors of constant term}}{\text{factor of leading coefficient}}$

These rational roots are possible rational roots not the actual roots of f(x).

For actual roots of f(x), the value of the function is 0.

Put each value of x from the option in the function, if we get the value of f(x) is 0, then that value is the actual root of f(x).

Put x=3.

$f(3)=60(3)^2-57(3)-18=351\neq 0$

Put x=6

$f(6)=60(6)^2-57(6)-18=1800\neq 0$

Put $x=-\frac{1}{4}$

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

Put $x=-\frac{6}{5}$

$f(-\frac{6}{5})=60(-\frac{6}{5})^2-57(-\frac{6}{5})-18=136.8\neq 0$

Only for $x=-\frac{1}{4}$ the value of function is 0, therefore the correct option is third.

Answer 2

Answer:

the answer is -1/4

Step-by-step explanation: