Question

How do you solve the inequality `x^2-6x-7<0`?

Answer 1

`-1< x <7`

Explanation:

First, factorise the x terms.

`x^2-6x-7<0`

`(x-7)(x+1)<0`

To find the roots:

`x-7=0 or x+1=0`

`x=7 or x=-1`

The question asks us to find the portion of the curve which is less than zero `(<0`).
Thus, we need to show by giving a range of values of `x`.

Do a rough sketch of a graph showing
- `x^2` curve on x and y axis
- curve must be positive, meaning it has a minimum point. (like a smile)
- label the x-intercepts with the values found above.

This is a zoomed in graph
graph{x^2-6x-7 [-6.875, 13.125, -4.96, 5.04]}

This next step is a marking point.
Recall when drawing a graph, you have to mark an "x" or "+" at each coordinates. This is different when involved with inequalities.

• When the equality is: `< , >`, you have to draw a circle around the x-intercept, instead of marking an "x".

• When the equality is: `<= , >=`, you have to shade a dot at the x-intercept, instead of marking an "x".

Since the equality given is "<", draw a small circle at the intercepts.

The portion of the curve that is negative falls below `y=0`

This means the curve is negative between `x=-1` and `x=7`

Which is expressed as `color(red)(-1 < x <7)`

(If the question states `x^2-6x-7>0`, the portions of the curve that are positive is when `x<-1 and x>7`)

Answer 2

`-1 < x< 7`

Explanation:

`"factorise the quadratic on the left side"`

`rArr(x-7)(x+1)<0`

`"find the zeros"`

`x=-1" and " x=7`

`"these indicate where the function changes sign"`

`"the zeros 'split' the x-axis into 3 intervals"`

`x < -1,color(white)(x)-1 < x<7,color(white)(x)x>7`

`"consider a "color(blue)"test point "" in each interval"`

`"we want to find where the function is negative ", <0`

`"substitute each test point into the function and "`
`"consider it's sign"`

`color(red)(x=-2)to(-)(-)tocolor(red)" positive"`

`color(red)(x=2)to(-)(+)tocolor(blue)" negative"`

`color(red)(x=10)to(+)(+)tocolor(red)" positive"`

`rArr-1 < x < 7" is the solution"`