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

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#)

#"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"#