Question

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

Answer 1

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 -6x+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^2-6x+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 `+//-` signs, depending on where each factor is positive or negative. For instance, the top factor `x-(3-sqrt2)` 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 `(-)xx(-) =(+)`:

`[(,|,,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^2-6x+7` is less than 0: between `3-sqrt2` and `3+sqrt2`, not including these endpoints.

This is written in interval notation as

`x in (3-sqrt2,"  "3+sqrt2)`.