Question

Yuri thinks 3/4 is a root of the following function.

q(x)=6x^3+19x^2-15x-28
Explain to Yuri why 3/4 cannot be a root.

Answer 1

Since
`q((3)/(4)) \\eq 0`, 3/4 cannot be a root of q(x).

Explanation:

Root of a function:

If
`x^(\ast)` is a root of a function f(x), we have that
`f(x^(\ast)) = 0`.

In this question.

We have to find
`q((3)/(4))`. So

`q((3)/(4)) = 6((3)/(4))^3 + 19((3)/(4))^2 - 15((3)/(4)) - 28`

`q((3)/(4)) = (162)/(64) + (171)/(16) - (45)/(4) - 28`

`q((3)/(4)) = (162)/(64) + (684)/(64) - (720)/(64) - (1792)/(64)`

`q((3)/(4)) \\eq 0`

Since
`q((3)/(4)) \\eq 0`, 3/4 cannot be a root of q(x).