Question

If f(x)=x squared-2x and g(x) = 6x+4 for which is a value of x does (f+g)(x) =0

Answer 1

Answer:

For $x = -2$

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

In this question:

$f(x) = x^2 - 2x$

$g(x) = 6x + 4$

$f(x) + g(x) = (f+g)(x) = x^2 - 2x + 6x + 4 = x^2 + 4x + 4$

Which is a value of x does (f+g)(x) =0

$x^2 + 4x + 4$

Quadratic equation with $a = 1, b = 4, c = 4$

So

$\Delta = 4^{2} - 4*1*4 = 0$

$x_{1} = \frac{-4 + \sqrt{0}}{2} = -2$

$x_{2} = \frac{-4 - \sqrt{0}}{2} = -2$

For $x = -2$