Bhaskara (circa 6th century AD)

1. Grade 9 - Mathematics
Topic 1.8 :
Solving Quadratic
Inequalities

2. OBJECTIVES :
•Define the interval
correctly
•Express the Quadratic
Inequalities
•Solve the Quadratic
Inequalities through graphs
and interval notation

3. Review: Solve the Linear Inequalities:
1.x2 6x + 8 > 0−
2. x2 + 2x +1 ≥ 0
3. x2 + x +1 > 0
4.7x2 + 21x 28 < 0−
5. x2 + 4x 7 < 0− −
6. 4x2 – 16 > 0
7. 4x2 4x + 1 ≤ 0−
8. X4+ 12x3 – 64x2 > 0
9. x4 25x2 + 144 < 0−
10. x4 16x2 225 ≥ 0− −

4. • Discussion of the lesson: QUADRATIC
INEQUALITIES are inequalities which can
be written in one of the following forms:
ax2 +bx2+c> 0
ax2 +bx2+c<0
ax2 +bx2+c>0
ax2 +bx2+c<0
where a,b and c are real numbers.

5. • EXAMPLE NO.1: Solve x – 4 < 0
To solve this inequality, all I have to do is
add the 4 to the other side to get the
solution "x < 4". So I already know what
the answer is. But now I'll approach this
problem from a different angle, by
considering the related two-variable graph.
For "x – 4 < 0", the associated two-variable
linear graph is y = x – 4:

6. • The inequality "x – 4 < 0"is asking "when is
the line y = x – 4 below the line y = 0?"Since
the line y = 0 is just the x-axis, the inequality
is therefore asking "when is the line y = x – 4
below the x-axis?"The first step in answering
this question is to find where the line crosses
the x-axis; that is, first I need to find the x-
intercept. So I set y equal to zero and solve:
y = x – 4
0 = x – 4
4 = x

7. • So the line y = x – 4 crosses the x-axis at
x = 4. Since the line y = x – 4 is a straight
line, it will be above the x-axis on one
side of the intercept and below the x-axis
on the other side of the intercept.

8. • Since the slope of this line is m = 1 (in particular,
since the slope is positive), then the line is
increasing, so the line is below the axis on the
left-hand side (before the intercept) and
above the axis on the right-hand side (after
the intercept), as is highlighted at right:

9. • The original question asked me to
solve x – 4 < 0, so I need to find where
the line is below the x-axis. This
happens on the left-hand side of the
intercept:

11. • That is, by looking at the graph of the
associated line and determining where (on
the x-axis) the graphed line was below the
x-axis, you can easily see that the solution
to the inequality "x – 4 < 0" is the
inequality "x < 4". You can follow the same
method of finding intercepts and using
graphs to solve inequalities containing
quadratics.

12. EXAMPLE NO. 2
Solve 2x2 + 4x >
x2 – x – 6.

14. • As you can see, it is hard to tell where the
green line (y = 2x2 + 4x) is above the blue
line (y = x2 – x – 6). So, instead of trying
to solve this inequality, I will instead work
with the following related inequality:
• 2x2 + 4x > x2 – x – 6
• 2x2 + 4x – x2 + x + 6 > 0
• x2 + 5x + 6 > 0

15. • This last inequality is simpler to deal with because
now all I have to do is find the zeroes of y =x2 + 5x +
6 (which is easy) and then pick the correct intervals
based on just the one parabola (which is also easy).
That is, it is simpler to compare one parabola with
the x-axis than to compare two parabolas with
each other. But since the one parabola (y = x2 + 5x
+ 6) came from combining the two original
parabolas ("paraboli"?), the solution to the simpler
one-parabola inequality will be the same as the
solution to the original two-parabola inequality.
Since the solutions will be the same, I'll work
with the simpler case.

17. • These two intercepts split the number-line
into three intervals, namely x < –3, –3 < x <
–2, and x > –2. On which of these three
intervals is y = x2 + 5x + 6above the x-axis?
Since y = x2 + 5x + 6graphs as a right-side-
up parabola, the quadratic is above the axis
on the ends:
• Then the solution is:
• x < –3 or x > –2

18. • Why was this solution "or equal to", rather than
just "greater than" or "less than"? Because the
original inequality was "or equal to", so the
boundary points, being the zeroes or x-intercepts,
are included in the solution.
• The above solution could also be stated as "all
real numbers" or written as the interval "from
negative infinity to positive infinity".
• There is one fiddly case that you might not even
have to deal with, but I'll cover it anyway, just in
case your teacher likes tricky test problems.

19. • The above solution could also be stated
as "all real numbers" or written as the
interval "from negative infinity to
positive infinity".
• There is one fiddly case that you might
not even have to deal with, but I'll cover
it anyway, just in case your teacher
likes tricky test problems

20. EXAMPLE NO. 3
Solve –x2 + 6x – 9 > 0.

21. • First, I'll find the zeroes of y = –x2 + 6x – 9, the
associated quadratic equation:
–x2 + 6x – 9 = 0
x2 – 6x + 9 = 0
(x – 3)(x – 3) = 0
x = 3

22. • I need to find where y = –x2 + 6x – 9 is above the axis.
But I know (and can verify from the above graph) that
this quadratic only touches the axis from below; it is
never fully above the axis. However, this inequality is an
"or equal to" inequality, so the "equal" part counts as
part of the solution. That is, the intercept is part of the
solution. In this case, it is actually the only solution,
because the graph only touches the axis (is equal to
zero); it never goes above (is never greater than zero).
So the solution is
x = 3

23. EXAMPLE NO. 4
Solve –x2 + 6x – 9 > 0

24. • In this case, they're asking me for where the
graph is strictly above the axis. Since the graph
only touches the axis from below, and never
crosses, then the graph is never above the axis,
so there is:
NO SOLUTION.
• If you are careful about finding the zeroes of
the quadratic, and use your knowledge of the
shape of quadratic graphs, you shouldn't have
any trouble solving quadratic inequalities.

25. • Practice Yourself: SOLVE THE FOLLOWING AND
GRAPH THE CORRECT ANSWER
1. y 2 −17y + 70 < 0
2. x 2 + 9x + 13 > −7
3. x(x +1) >112 − 5x
4. a 2 + 3a + 2 < −3(a + 2)
5. 2x 2 ≤ 5x − 2
6. 10 − 9y ≥ −2y 2
7. b(b + 3) ≥ −2
8. a 2 ≤ 4(2a − 3)
9. y 2 −17y + 70 < 0
10. x 2 + 9x + 13 > −7

26. Challenge Yourself:
• Although there is not a single known inventor of
the quadratic formula, its use dates back to the
Middle Kingdom in Egypt. Greek mathematicians,
such as Euclid also used early versions of the
quadratic formula but there is another one who
also used it. Match the following to identify who it is
by solving the following:

29. Answers in practice yourself:
1.S = (-∞, 2) (4, ∞)
2.(x + 1)2 ≥ 0
As a number squared is always positive.
S =
3.P(0) = 0 + 0 + 1 > 0
The sign obtained coincides with the inequality,
the solution is.
4.
(−4, 1)
5.P(0) = −02 + 4 •0 − 7 < 0
S =

30. 10

31. 10.
(-∞, −5] [5, +∞)
Answers in Challenge yourself:
1. x < -14 or x > 8
2. no solution
3. −4 ≤ d ≤ 3 2
4. x ≤ -4 or x ≥ -2
5. y ≤ 2 or y ≥ 5 2
6. 1 2 ≤ x ≤ 2
7. -6 < c < 5
8. a < -5 or a > 1 3
9. b ≤ -2 or b ≥ -1
10. 2 < a < 6