Question

If f(x) = x2 - 2x and g(x) = 6x + 4 for which value of x does (t +g)(x)=0​

Answer 1

Answer:

-2

Assumption:

Find the value of x such that $(f+g)(x)=0$.

Step-by-step explanation:

$(f+g)(x)=0$

$f(x)+g(x)=0$

$(x^2-2x)+(6x+4)=0$

Combine like terms:

$x^2+4x+4=0$

This is not too bad too factor on the left hand side since 2(2)=4 and 2+2=4.

$(x+2)(x+2)=0$

$(x+2)^2=0$

So we need to solve:

$x+2=0$

Subtract 2 on both sides:

$x=-2$

Let's check:

$(f+g)(-2)$

$f(-2)+g(-2)$

$((-2)^2-2(-2))+(6(-2)+4)$

$(4+4)+(-12+4)$

$(8)+(-8)$

$0$

0 was the desired output of $(f+g)(x)$.