The function gx is defined as gx=6x^2+23x-4,when does g(x)=0

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Riia Riia · Answer 1 · 2018-09-04T07:39:03+02:00

The given function is

[tex] g(x) = 6x^2 +23x-4 [/tex]

And we have to find the value of g(0), and for that we put 0 for x in the given equation.

And on substituting 0 for x in the given expression, we will get

[tex] g(0) = 6(0)^2 +23(0) -4 [/tex]

[tex] g(0) = 0 + 0 -4 = -4 [/tex]

So for the given function, the value of g(0) is -4 .

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carlosego carlosego · Answer 2 · 2018-09-04T11:30:25+02:00

For this case we have the following function:

[tex] g (x) = 6x ^ 2 + 23x-4[/tex]

What we must do is find the zeros of the function.

We have then:

[tex] 6x ^ 2 + 23x-4 = 0[/tex]

Factoring we have:

[tex] (x + 4) (6x-1) = 0[/tex]

From here, we solve the solutions of the equation:

Solution 1:

[tex] x + 4 = 0 x = -4 [/tex]

Solution 2:

[tex] 6x-1 = 0[/tex]

[tex] x = \frac{1}{6}[/tex]

Answer:

The values of x that make the equation equal to zero are:

[tex] x = -4 [/tex]

[tex] x = \frac{1}{6}[/tex]