# How do you solve the inequality and express your answer in interval notation; x ²-6x+7<0?

Aug 5, 2017

Set the polynomial equal to 0 to find its factors & roots; set up a sign-table to show the sign of each factor around all roots; multiply the rows in the table to arrive at the final row, which shows the sign of the whole polynomial.

#### Explanation:

Since ${x}^{2} - 6 x + 7$ is a polynomial, we know it is continuous everywhere. This means its roots are the boundary points of where it is above or below 0.

We set the polynomial equal to 0 to find its roots:

${x}^{2} - 6 x + 7 = 0$

Using the quadratic formula, we arrive at

x=("-"("-"6)+-sqrt(("-"6)^2-4(1)(7)))/(2(1))

$\textcolor{w h i t e}{x} = \frac{6 \pm \sqrt{36 - 28}}{2}$

$\textcolor{w h i t e}{x} = \frac{6 \pm \sqrt{8}}{2} \text{ } = 3 \pm \sqrt{2}$

Now, we set up a table like this one, with the factors of the polynomial on the left, and the roots of the polynomial at the top:

[(,|,,3-sqrt2,,3+sqrt2, ),(—————,+,—,——,—,——,—),(x-(3-sqrt2),|, , , , , ),(x-(3+sqrt2),|, , , , , ),(—————,+,—,——,—,——,—),(,|, , , , , )]

We then fill the table with $+ / -$ signs, depending on where each factor is positive or negative. For instance, the top factor $x - \left(3 - \sqrt{2}\right)$ is negative when $x$ is less than $3 - \sqrt{2}$, and positive when $x$ is greater:

[(,|,,3-sqrt2,,3+sqrt2, ),(—————,+,—,——,—,——,—),(x-(3-sqrt2),|,color(orange)-,color(orange)circ,color(orange)+,color(orange)+,color(orange)+),(x-(3+sqrt2),|,-,-,-,circ,+),(—————,+,—,——,—,——,—),(,|, , , , , )]

The bottom row gets filled with the product of the rows above it. For example, in the first column, we'll get $\left(-\right) \times \left(-\right) = \left(+\right)$:

[(,|,,3-sqrt2,,3+sqrt2, ),(—————,+,—,——,—,——,—),(x-(3-sqrt2),|,color(orange)-,circ,+,+,+),(x-(3+sqrt2),|,color(orange)-,-,-,circ,+),(—————,+,—,——,—,——,—),(x^2-6x+7,|,color(orange)+,0,-,0,+)]

This tells us exactly where the polynomial ${x}^{2} - 6 x + 7$ is less than 0: between $3 - \sqrt{2}$ and $3 + \sqrt{2}$, not including these endpoints.

This is written in interval notation as

$x \in \left(3 - \sqrt{2} , \text{ } 3 + \sqrt{2}\right)$.