Question

1. function ( g ) is defined by ( g(x) = 3x^2 - 2x - 5 ). what are the solutions to ( g(x) = 0 )?

options:
( x = 1 ) and ( x = -\frac{5}{3} )
( x = 1 ) and ( x = -\frac{3}{5} )
( x = -1 ) and ( x = \frac{5}{3} )
( x = -1 ) and ( x = \frac{3}{5} )

Explanation:

Step1: Set function equal to 0

$$3x^2 - 2x - 5 = 0$$

Step2: Use quadratic formula

For $ax^2+bx+c=0$, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=3$, $b=-2$, $c=-5$.
$$x=\frac{-(-2)\pm\sqrt{(-2)^2-4(3)(-5)}}{2(3)}$$

Step3: Simplify discriminant

$$\sqrt{4 + 60}=\sqrt{64}=8$$

Step4: Calculate two solutions

$$x=\frac{2+8}{6}=\frac{10}{6}=\frac{5}{3}$$
$$x=\frac{2-8}{6}=\frac{-6}{6}=-1$$

Answer:

$x=-1$ and $x=\frac{5}{3}$ (matches the third option: $x=-1$ and $x=\frac{5}{3}$)