# What is the domain of (x-7)/(x^2+3x+8)? I can't figure it out, thanks.

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Should not be 41, I placed (-8) for some reason. Should be -23. Which is not a real solution. Domain would be all real #'s

The domain of a function is the set of all possible inputs, or values for which x can be.

Since your function: `(x - 7) / (x^2+3x+8)`

has a denominator, the denominator can never equal zero. Therefore we want to find the values at which `x^2 + 3x + 8 = 0.`

Since we can't factor, we could use quadratic formula or complete the square to solve for x.

Quadratic formula: `(-3+-sqrt(3^2-4*1**8))/2`

Next, in simplifing we get: `(-3+-sqrt(41))/2`

**So, domain is all real #'s except `(-3+-sqrt(41))/2` **

**or domain is: `x!= (-3+-sqrt(41))/2` **

**or `x!= 1.702 or x!= -4.702` **

Let

`f(x)=(x-7)/(x^2+3x+8)` .Thus domain of function f(x) is the value of all x where f(x) is defined.

F(x) is not defined if `x^2+3x+8=0`

i.e. `x=(-3+-sqrt(3^2-4xx1xx8))/2`

`=(-3+_sqrt(-23))/2=(-3+-sqrt(23)i)/2` which is not real numbers.

the graph of g(x)=`x^2+3x+8` does not intersect x-axis .It shows that f(x) is defined for real numbers.

**Thus domain of function f(x) is all real numbers if f(x) is real function.**