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

##### 1 Answer

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

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

#x^2-6x+7=0#

Using the quadratic formula, we arrive at

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

#color(white)x=(6+-sqrt(36-28))/2#

#color(white)x=(6+-sqrt(8))/2" "=3+-sqrt2#

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

#[(,|,,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

#[(,|,,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

This is written in interval notation as

#x in (3-sqrt2," "3+sqrt2)# .