. If f(x)=x^2-2x and g(x)=6x+4 for which value of x does (f+g)(x)=0 -4 -2 2 4

To find the value of x for which $(f+g)(x)=0$, we need to add the functions f(x) and g(x) and then set the sum equal to 0.<br /><br />Given:<br />$f(x) = x^2 - 2x$<br />$g(x) = 6x + 4$<br /><br />$(f+g)(x) = f(x) + g(x)$<br />$(f+g)(x)2 - 2x) + (6x + 4)$<br />$(f+g)(x) = x^2 + 4x + 4$<br /><br />Now, we set $(f+g)(x)$ equal to 0 and solve for x:<br />$x^2 + 4x + 4 = 0$<br /><br />To solve this quadratic equation, we can use the quadratic formula:<br />$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />In this case, $a = 1$, $b = 4$, and $c = 4$.<br /><br />$x = \frac{-4 \pm \sqrt{4^2 - 4(1)(4)}}{2(1)}$<br />$x = \frac{-4 \pm \sqrt{16 - 16}}{2}$<br />$x = \frac{-4 \pm \sqrt}}{2}$<br />$x = \frac{-4}{2}$<br />$x = -2$<br /><br />Therefore, the value of x for which $(f+g)(x)=0$ is $x = -2$.