### Instruction: If $x$ is a real number and $x^2-7x+6<0$, what are the possible values for $x$? Use interval notation to express your answer.### Response: I need to find the values of $x$ that make the quadratic expression $x^2-7x+6$ negative.
One way to do this is to factor the quadratic and look at the signs of the factors.
I notice that $x^2-7x+6$ can be factored as $(x-1)(x-6)$.
Now I can use a sign chart to analyze the factors and their product.
I mark the zeros of the factors on a number line: 1 and 6.
I pick a test value in each interval and plug it into the factors.
For example, in the interval $(-\infty,1)$, I pick $x=0$ and see that $(0-1)(0-6) = 6 > 0$, so the product is positive in this interval.
I repeat this for the other intervals: $(1,6)$ and $(6,\infty)$.
I find that $(2-1)(2-6) = -4 < 0$, so the product is negative in this interval.
And $(7-1)(7-6) = 6 > 0$, so the product is positive in this interval.
This means that the product is negative only when $x$ is between 1 and 6, not including the endpoints.