1. x2 – 2x – 3
x2 + x – 2
In factored form =
(x – 3)(x + 1)
(x – 1)(x + 2)
So for x = –3/2:
(x – 3)(x + 1)
(x – 1)(x + 2)
=
(–)(–)
(–)(+)
< 0
For x = –1/2:
(x – 3)(x + 1)
(x – 1)(x + 2)
=
(–)(+)
(–)(+)
> 0
This leads to the sign charts of formulas. The sign–
chart of a formula gives the signs of the outputs.
Sign–Charts and Inequalities I
Example B. Determine whether the outcome is
x2 – 2x – 3
x2 + x – 2
if x = –3/2, –1/2.+ or – for
For polynomials or rational expressions,
factor them to determine the signs of their outputs.
2. Example C. Let f = x2 – 3x – 4 , use the sign–
chart to indicate when is f = 0, f > 0, and f < 0.
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0 x = 4 , –1
Mark off these points on a line:
(x–4)(x+1) + + + + + – – – – – + + + + +
0 4–1
Select points to sample in each segment:
Test x = – 2,
get – * – = + .
Hence the segment
is positive. Draw +
sign over it.
–2
Test x = 0,
get – * + = –.
Hence this segment
is negative.
Put – over it.
Test x = 5,
get + * + = +.
Hence this segment
is positive.
Put + over it.
5
Sign–Charts and Inequalities I
3. Example D. Make the sign chart of f =
Select a point to sample in each segment:
Test x = –3,
we've a
(x – 3)
(x – 1)(x + 2)
The root for f = 0 is from the zero of the numerator
which is x = 3. The zeroes of the denominator
x = 1, –2 are the values where f is undefined (UDF).
Mark these values on a real line.
(x – 3)
(x – 1)(x + 2) –2 1 3
UDF UDF f=0
–3
( – )
( – )( – )
= –
segment.
0 2 4
Test x = 0,
we've a
( – )
( – )( + )
= +
segment.
Test x = 2,
we've a
( – )
( + )( + )
segment.
= –
Test x = 4,
we've a
( + )
( + )( + )
segment.
= +
– – – – + + + – – – + + + +
Sign–Charts and Inequalities I
4. Example E. Solve x2 – 3x > 4
0 4–1
The solutions are the + regions: (–∞, –1) U (4, ∞)
–2 5
4–1
Note: The empty dot means those numbers are excluded.
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart.
Draw the sign–chart, sample the points x = –2, 0, 5
(x – 4)(x + 1)
+ + + – – – – – – + + + +
Setting one side to 0, we have x2 – 3x – 4 > 0 or
(x – 4)(x + 1) > 0. The roots are x = –1, 4.
Sign–Charts and Inequalities I
5. Example F. Solve x – 2
2 <
x – 1
3
Set the inequality to 0, x – 2
2
x – 1
3
< 0
Put the expression into factored form,
x – 2
2
x – 1
3
=
(x – 2)(x – 1)
2(x – 1) – 3(x – 2)
=
(x – 2)(x – 1)
– x + 4
Hence the inequality is (x – 2)(x – 1)
– x + 4 < 0
Draw the sign chart by sampling x = 0, 3/2, 3, 5
It has a root at x = 4, and it's undefined at x = 1, 2.
410 5
+ + + – – + + + + – – – –
23/2 3
UDF UDF
(x – 2)(x – 1)
– x + 4
The answer are the shaded negative regions,
i.e. (1, 2) U [4 ∞).
Sign–Charts and Inequalities I