How do you solve the inequality #x^2-6x-7<0#?
2 Answers
Explanation:
First, factorise the x terms.
To find the roots:
The question asks us to find the portion of the curve which is less than zero
Thus, we need to show by giving a range of values of
Do a rough sketch of a graph showing
-
- 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
This means the curve is negative between
Which is expressed as
(If the question states
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"#