Transcript 4.9 SOLVING QUADRATIC INEQUALITIES EXAMPLE 4 Solve a quadratic inequality using a table Solve x2 + x ≤ 6 using a table. SOLUTION Rewrite the inequality.

4.9 SOLVING
QUADRATIC
INEQUALITIES
EXAMPLE 4
Solve a quadratic inequality using a table
Solve x2 + x ≤ 6 using a table.
SOLUTION
Rewrite the inequality as x2 + x – 6 ≤ 0.
Then make a table of values.
Notice that x2 + x – 6 ≤ 0 when the values of x are
between –3 and 2, inclusive.
ANSWER
The solution of the inequality is –3 ≤ x ≤ 2.
EXAMPLE 5
Solve a quadratic inequality by graphing
Solve 2x2 + x – 4 ≥ 0 by graphing.
SOLUTION
The solution consists of the x-values for which the
graph of y = 2x2 + x – 4 lies on or above the x-axis. Find
the graph’s x-intercepts by letting y = 0 and using the
quadratic formula to solve for x.
0 = 2x2 + x – 4
x = –1+ 12– 4(2)(–4)
2(2)
–1+ 33
x= 4
x 1.19 or x
–1.69
EXAMPLE 5
Solve a quadratic inequality by graphing
Sketch a parabola that opens
up and has 1.19 and –1.69 as
x-intercepts. The graph lies
on or above the x-axis to the
left of (and including)
x = –1.69 and to the right of
(and including) x = 1.19.
ANSWER
The solution of the inequality is approximately
x ≤ –1.69 or x ≥ 1.19.
GUIDED PRACTICE
5.
for Examples 4 and 5
Solve the inequality 2x2 + 2x ≤ 3 using a table and
using a graph.
ANSWER
–1.8 ≤ x ≤ 0.82
ALGEBRAICALLY
 x 2  12 x  32
 x  12 x  32  0
2
x 2  12 x  32  0
 x  4 x  8  0
x  4, x  8
-9 -8 -6 -4
  9   12 9   32
 81  108  32
27  32
2
true
0
  6   12 6   32
 36  72  32
2
36  32
false
 0 2  120  32
0  32
true
x  8, x  4
EXAMPLE 7
Solve a quadratic inequality algebraically
Solve x2 – 2x > 15 algebraically.
SOLUTION
First, write and solve the equation obtained by
replacing > with = .
x2 – 2x = 15
x2 – 2x – 15 = 0
(x + 3)(x – 5) = 0
x = –3 or x = 5
Write equation that corresponds
to original inequality.
Write in standard form.
Factor.
Zero product property
EXAMPLE 7
Solve a quadratic inequality algebraically
The numbers –3 and 5 are the critical x-values of the
inequality x2 – 2x > 15. Plot –3 and 5 on a number line,
using open dots because the values do not satisfy the
inequality. The critical x-values partition the number
line into three intervals. Test an x-value in each
interval to see if it satisfies the inequality.
Test x = – 4:
(–4)2 – 2(–4) = 24 > 15
Test x = 1:
12 – 2(1)= –1 >15
ANSWER
The solution is x < –3 or x > 5.
Test x = 6:
62 –2(6) = 24 >15
Solve the inequality using any method.