Solve Rational Inequalities - More Examples

More examples on solving rational inequalities.


Example 4 Solve the rational inequality given by

\dfrac{2x + 1}{x+2} \ge \dfrac{x-4}{x - 3}

Solution to Example 4

Given Subtract (x - 4) / (x - 3) from both sides of the given inequality so that its right side equal to zero.

\dfrac{2x + 1}{x+2} - \dfrac{x-4}{x - 3} \ge 0

Rewrite the two rational expressions on the left side with common denominator.

inequality with common denominator question 4

Add the two rational expressions and simplify the numerator to obtain

\dfrac{x^2- 3x + 5}{(x+2)(x-3)} \ge 0

Zeros of numerator and denominator
The numerator is a quadratic expression of the form ax2 + bx + c whose discriminant , b2 - 4(a)(c) = (-3)2 - 4(1)(5) = -11 , is negative and therefore has no zeros.
The zeros of the denominator: (x + 2)(x - 3) = 0 are x = -2 and x = 3.

The two zeros divide the real number line into 3 intervals as follows:

(- , -2) , (-2 , 3) and (3, +)

Select test values that are within each of the 3 intervals above and test the rational expression

\dfrac{x^2- 3x + 5}{(x+2)(x-3)}
to find its sign within each interval.
a) interval (- , -2)
test value x = - 3
We now evaluate
\dfrac{x^2- 3x + 5}{(x+2)(x-3)}
at x = -3 to find its sign.

\dfrac{(-3)^2- 3(-3) + 5}{((-3)+2)((-3)- 3)} = \dfrac{23}{6}
which is positive .
b) interval (-2 , 3)
test value x = 0
We now evaluate
\dfrac{x^2- 3x + 5}{(x+2)(x-3)}
at x = 0 to find its sign.

\dfrac{(0)^2- 3(0) + 5}{((0)+2)((0)- 3)} = -\dfrac{5}{6}
which is negative .
c) interval (3 , )
test value x = 4
We now evaluate
\dfrac{x^2- 3x + 5}{(x+2)(x-3)}
at x = 4 to find its sign.

\dfrac{(4)^2- 3(4) + 5}{((4)+2)((4)- 3)} = \dfrac{9}{6}
which is positive .

We now put all the above in a table of sign.

table of sign for question 4

Conclusion The solution set of the given rational inequality is given by the interval

(- , -2) U (3 , )

Graphical solution to the inequality.
In the first step above, we obtained the inequality

\dfrac{2x + 1}{x+2} - \dfrac{x-4}{x - 3} \ge 0
which is equivalent to the given inequality. Below is shown the graph to the function
y = \dfrac{2x + 1}{x+2} - \dfrac{x-4}{x - 3}
which represents the left side of the inequality. It is easy to check graphically that y is positive over the interval
(- , -2) U (3 , )
.
graphical solution to the inequality in question 4


Example 5 Solve the rational inequality given by

\dfrac{x}{x^2 - 2x - 3} \le \dfrac{1}{x + 3}

Solution to Example 5

Subtract 1 / (x + 3) from both sides of the inequality so that its right side equal to zero.

\dfrac{x}{x^2 - 2x - 3} - \dfrac{1}{x + 3} \le 0

Factor the quadratic expression x2 - 2 x - 3 in the denominator.

\dfrac{x}{(x-3)(x+1)} - \dfrac{1}{x + 3} \le 0

Rewrite the rational the two expressions on the left side with the common denominator (x - 3)(x + 1)(x + 3) .

inequality with common denominator question 5

Add and simplify the two rational expressions on the left side of the inequality and simplify the numerator to obtain

\dfrac{5x + 3}{(x-3)(x+1)(x+3)} \le 0

Zeros of numerator and denominator
The zero of the numerator: 5x + 3 = 0, x = -3/5.
The zeros of the denominator: (x - 3)(x + 1)(x + 3) = 0 are x = 3, x = -1 and x = -3.

The four zeros divide the real number line into 5 intervals as follows:

(- , -3) , (-3 , -1) , (-1 , -3/5) , (-3/5 , 3) , (3 , +)

Select and test values that are within each interval and test the rational expression

\dfrac{x}{x^2 - 2x - 3} - \dfrac{1}{x + 3}
and find its sign.
a) interval (- , -3)
test value x = -4
We now evaluate
\dfrac{x}{x^2 - 2x - 3} - \dfrac{1}{x + 3}
at x = -4 to find its sign.

\dfrac{(-4)}{(-4)^2 - 2(-4) - 3} - \dfrac{1}{(-4) + 3} = \dfrac{17}{21}
which is positive

b) interval (-3 , -1)
test value x = -2
We now evaluate

\dfrac{x}{x^2 - 2x - 3} - \dfrac{1}{x + 3}
at x = -2
\dfrac{(-2)}{(-2)^2 - 2(-2) - 3} - \dfrac{1}{(-2) + 3} = - \dfrac{7}{5}
which is negative

c) interval (-1 , -3/5)
test value x = -0.8
We now evaluate

\dfrac{x}{x^2 - 2x - 3} - \dfrac{1}{x + 3}
at x = -0.8 to find its sign.
\dfrac{(-0.8)}{(-0.8)^2 - 2(-0.8) - 3} - \dfrac{1}{(-0.8) + 3} \approx 0.6
which is positive

d) interval (-3/5 , 3)
test value x = 0
We now evaluate

\dfrac{x}{x^2 - 2x - 3} - \dfrac{1}{x + 3}
at x = 0 to find its sign.
\dfrac{0}{0^2 - 2(0) - 3} - \dfrac{1}{(0)+ 3} = - \dfrac{1}{3}
which is negative

d) interval (3 , + )
test value x = 4
We now evaluate

\dfrac{x}{x^2 - 2x - 3} - \dfrac{1}{x + 3}
at x = 4 to find its sign.
\dfrac{(4)}{(4)^2 - 2(4) - 3} - \dfrac{1}{(4) + 3} = \dfrac{23}{35}
which is positive.

Let us now put all the above results in a table of sign

table of sign for question 5

Conclusion The solution set of the given rational inequality is given by the interval

(-3 , -1) U [ -3/5 , 3 )

Graphical solution to the inequality.
We obtained above an equivalent (to the given) inequality to solve , which is

\dfrac{x}{x^2 - 2x - 3} - \dfrac{1}{x + 3} \le 0

Below is shown the graph to the function
y = \dfrac{x}{x^2 - 2x - 3} - \dfrac{1}{x + 3}
which represents the left side of the inequality. It is easy to check graphically that y is negative or zero over the interval
(-3 , -1) U [ -3/5 , 3 )
.
graphical solution to the inequality in question 5


Example 6  Solve the rational inequality given by

\dfrac{3x^2}{|x^2-3x-4|} \ge \dfrac{1}{3}

Solution to Example 6

Write the inequality with the right side equal to zero by subtracting
1 / 3 from both sides.

\dfrac{3x^2}{|x^2-3x-4|} - \dfrac{1}{3} \ge 0

Rewrite the left side of the inequality so that all terms have common denominator.

inequality with common denominator question 6

Group

\dfrac{9x^2- |x^2-3x-4|} {3|x^2-3x-4|} \ge 0

The zeros of the numerator are found by solving two equations: (the expression within the absolute value symbol may be positive or negative)
1) 9x2 - (x2 - 3x - 4) = 0
8x2+ 3x + 4 = 0 , discriminant = 32 - 4(8)(4) < 0 , this equation has no real solutions
2) 9x2 - (-1)(x2 - 3x - 4) = 0
10x2 - 3x - 4 = 0 , two solutions: x = -1/2 and x = 4/5.

The zeros of the denominator: x2 - 3x - 4 = 0 are x = - 1 and x = 4

The four zeros divide the real number line into five intervals (Note that the zeros are ordered from the smallest to the largest).

(- , -1) , (-1 , -1/2) , ( -1/2 , 4/5) , ( 4/5 , 4) , (4 , +)

Select test values that are within each interval and use them to find the sign of the expression

\dfrac{3x^2}{|x^2-3x-4|} - \dfrac{1}{3}
which the right hand side of the inequality obtained above in step 1.
a) interval (- , -1)
test value x = - 2
We now evaluate
\dfrac{3x^2}{|x^2-3x-4|} - \dfrac{1}{3}
at x = - 2 to find its sign.

\dfrac{3( - 2)^2}{|( - 2)^2-3( - 2)-4|} - \dfrac{1}{3} = \dfrac{5}{3}
which is positive.

b) interval (-1 , -1/2)
test value x = - 0.75
We now evaluate

\dfrac{3x^2}{|x^2-3x-4|} - \dfrac{1}{3}
at x = - 0.75 to find its sign.
\dfrac{3(- 0.75)^2}{|(- 0.75)^2-3(- 0.75)-4|} - \dfrac{1}{3} \approx 1.09
which is positive.

c) interval (-1/2 , 4/5)
test value x = 0
We now evaluate

\dfrac{3x^2}{|x^2-3x-4|} - \dfrac{1}{3}
at x = 0 to find its sign.
\dfrac{3(0)^2}{|(0)^2-3(0)-4|} - \dfrac{1}{3} = - \dfrac{1}{3}
which is negative

d) interval (4/5 , 4)
test value x = 1
We now evaluate

\dfrac{3x^2}{|x^2-3x-4|} - \dfrac{1}{3}
at x = 1 to find its sign.
\dfrac{3(1)^2}{|(1)^2-3(1)-4|} - \dfrac{1}{3} = \dfrac{1}{6}
which is positive.

d) interval (4 , + )
test value x = 5
We now evaluate

\dfrac{3x^2}{|x^2-3x-4|} - \dfrac{1}{3}
at x = 5 to find its sign.
\dfrac{3(5)^2}{|(5)^2-3(5)-4|} - \dfrac{1}{3} = \dfrac{73}{6}
which is positive.



Let us now put all the above results in a table of sign.

table of sign for question 6

Conclusion The solution set of the given rational inequality is given by the interval
(- , -1) U (-1 , -1/2] U [ 4/5 , 4) U (4 , +)

Graphical solution to the inequality.

Shown below is an equivalent (to the given) inequality to solve which was obtained in step 1.

\dfrac{3x^2}{|x^2-3x-4|} - \dfrac{1}{3} \ge 0

Below is shown the graph to the function
y = \dfrac{3x^2}{|x^2-3x-4|} - \dfrac{1}{3}
which represents the left side of the inequality. It is easy to check graphically that y is positive or zero over the interval
(- , -1) U (-1 , -1/2] U [ 4/5 , 4) U (4 , +)
.
graphical solution to the inequality in question 6

More rational inequalities with detailed examples.

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