IXL - Solve quadratic inequalities (Precalculus practice)

Solve for

(

)

(

–

)

Write the solution using interval notation. Use the union symbol ∪ to express the solution as a union of disjoint intervals. Finite endpoints of all intervals should be integers. If there are no solutions, use the symbol ∅ for the empty set. Use the set notation

{

}

to represent an isolated solution

ref_doc_title.

Learn with an example

question

Solve for

(

–

)

(

–

)

≤

{

}

to represent an isolated solution

key idea

solution

Look at the polynomial inequality:

(

–

)

(

–

)

≤

Since the inequality is not strict (≤), the solution to the inequality is the set of values where the polynomial

(

–

)

(

–

)

is negative or zero.

Create a sign chart.

Start by marking the roots of the polynomial on a number line. You can find the roots by solving

(

–

)

(

–

)

The roots are 2 and 4.

The roots separate the number line into intervals. Find the sign of each factor on each interval.

The factor
x
–
2
is negative if
x
is in
(
–
∞
,
2
)
and positive if
x
is in
(
2
,
∞
)
.
The factor
x
–
4
is negative if
x
is in
(
–
∞
,
4
)
and positive if
x
is in
(
4
,
∞
)
.

Summarize this information in a table. Then, multiply the signs in each column to determine the sign of

(

–

)

(

–

)

on each of the intervals.

	( – ∞ , 2 )	( 2 , 4 )	( 4 , ∞ )
x – 2	-	+	+
x – 4	-	-	+
( x – 2 ) ( x – 4 )	+	-	+

The last row of the table shows that

(

–

)

(

–

)

is positive when

is in

(

–

∞

)

negative when

is in

(

)

and positive when

is in

(

∞

)

Write the solution set.

The polynomial is negative on

(

)

It is zero at 2 and 4 because those are its roots. So, it is negative or zero on

[

]

The solution set is:

[

]

Looking for more? Try this:

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P.1Graph solutions to quadratic inequalities

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