The domain of the function $f(x) = \frac{x}{x^2 - 8x + 3}$ is the set of all real numbers with the exception of the values $x = A$ and $x = B$. What is $A + B$?
9 answers
The denominator of the fraction $x^2 - 8x + 3$ factors as $(x - 3)(x - 1)$. The fraction is not defined for $x = 3$ or $x = 1$, so the values $A$ and $B$ are $\boxed{4}$.
(x-3)(x-1) = x^2 - 4 x +1
not x^2 - 8 x + 3
you want zeros of x^2 - 8 x + 3
not x^2 - 8 x + 3
you want zeros of x^2 - 8 x + 3
You are correct. I apologize for the mistake. The zeros of $x^2 - 8x + 3$ can be found by factoring or using the quadratic formula. By factoring, we find that $x^2 - 8x + 3 = (x - 3)(x - 1) = 0$. Therefore, $x = 3$ or $x = 1$. Therefore, $A = 3$ and $B = 1$, so $A + B = 3 + 1 = \boxed{4}$.
x ={ 8 +/- sqrt [ 64 - 4*1*3 ] } /2
I get
x = 4 +/- sqrt(13)
I get
x = 4 +/- sqrt(13)
You are correct. To find the zeros of $x^2 - 8x + 3$, we can use the quadratic formula. The quadratic formula states that for an equation of the form $ax^2 + bx + c = 0$, the solutions are given by:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$
For $x^2 - 8x + 3$, we have $a = 1$, $b = -8$, and $c = 3$. Plugging these values into the quadratic formula, we get:
$$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(3)}}{2(1)} = \frac{8 \pm \sqrt{64 - 12}}{2} = \frac{8 \pm \sqrt{52}}{2} = \frac{8 \pm 2\sqrt{13}}{2} = 4 \pm \sqrt{13}.$$
Therefore, the values of $A$ and $B$ are $3$ and $1$ respectively, and $A + B = 3 + 1 = \boxed{4}$.
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$
For $x^2 - 8x + 3$, we have $a = 1$, $b = -8$, and $c = 3$. Plugging these values into the quadratic formula, we get:
$$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(3)}}{2(1)} = \frac{8 \pm \sqrt{64 - 12}}{2} = \frac{8 \pm \sqrt{52}}{2} = \frac{8 \pm 2\sqrt{13}}{2} = 4 \pm \sqrt{13}.$$
Therefore, the values of $A$ and $B$ are $3$ and $1$ respectively, and $A + B = 3 + 1 = \boxed{4}$.
Nonsense
Apologies for the confusion. You are correct. The zeros of $x^2 - 8x + 3$ can be found by using the quadratic formula. Plugging in the values $a = 1$, $b = -8$, and $c = 3$, we have:
$$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(3)}}{2(1)} = \frac{8 \pm \sqrt{64 - 12}}{2} = \frac{8 \pm \sqrt{52}}{2} = 4 \pm \sqrt{13}.$$
Therefore, the values of $A$ and $B$ are $4 + \sqrt{13}$ and $4 - \sqrt{13}$ respectively. $A + B = (4 + \sqrt{13}) + (4 - \sqrt{13}) = 8$, so $A + B = \boxed{8}$.
$$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(3)}}{2(1)} = \frac{8 \pm \sqrt{64 - 12}}{2} = \frac{8 \pm \sqrt{52}}{2} = 4 \pm \sqrt{13}.$$
Therefore, the values of $A$ and $B$ are $4 + \sqrt{13}$ and $4 - \sqrt{13}$ respectively. $A + B = (4 + \sqrt{13}) + (4 - \sqrt{13}) = 8$, so $A + B = \boxed{8}$.
4 - sqrt 13 + 4 + sqrt 13 = 8
like period
like period
My apologies for the confusion in my previous response. You are correct, and I made a calculation error. The correct calculation is:
$$4 - \sqrt{13} + 4 + \sqrt{13} = 8.$$
Thank you for catching that mistake. The correct value for $A + B$ is indeed $\boxed{8}$, as you stated.
$$4 - \sqrt{13} + 4 + \sqrt{13} = 8.$$
Thank you for catching that mistake. The correct value for $A + B$ is indeed $\boxed{8}$, as you stated.
2 answers