Solve Rational Inequalities - More Examples

More examples on solving rational inequalities.


Example 4 Solve the rational inequality given by


Solution to Example 4

Given

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

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

Add the two rational expressions and simplify the numerator to obtain

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 interval and test the rational expression to find its sign.

a) interval (- , -2)

test value x = -3

We now evaluate at x = -3 to find its sign.

(positive).

b) interval (-2 , 3)

test value x = 0

We now evaluate at x = 5 to find its sign.

(negative).

c) interval (3 , )

test value x = 4

We now evaluate at x = 5 to find its sign.

(positive).

We now put all the above in a table.

-∞

-2   3
+ undefined - undefined +

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



which is equaivalent to the given inequality. Below is shown the graph to the function y = (2x + 1)/(x + 2) - (x - 4)/(x - 3) which represents the left side of the inequality. It is easy to check graphically that y = (2x + 1)/(x + 2) - (x - 4)/(x - 3) is positive over the interval
(- , -2) U (3 , )
.

graphical solution to the inequality in question 4



Example 5 Solve the rational inequality given by

Solution to Example 5

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

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

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

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

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 to find its sign.

a) interval (- , -3)

test value x = -4

We now evaluate at x = -4 to find its sign.

(positive)

b) interval (-3 , -1)

test value x = -2

We now evaluate at x = -2 to find its sign.

(negative)

c) interval (-1 , -3/5)

test value x = -0.8

We now evaluate at x = -0.8 to find its sign.

(positive)

d) interval (-3/5 , 3)

test value x = 0

We now evaluate at x = 0 to find its sign.

(negative)

d) interval (3 , + )

test value x = 4

We now evaluate at x = 4 to find its sign.

(positive)

Let us now put all the above results in a table

-∞

-3   -1   -3/5   3

+

undefined - undefined + 0 - undefined +

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



Below is shown the graph to the function y = x / (x
2 -2x - 3) - 1/(x + 3) which represents the left side of the inequality. It is easy to check graphically that y = x / (x2 -2x - 3) - 1/(x + 3) 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


Solution to Example 3

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

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

Group

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 , discriminat = 3x2-4(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 which the right hand side of the inequality obtained above in step 1.

a) interval (- , -1)

test value x = - 2

We now evaluate at x = - 2 to find its sign.

(positive)

b) interval (-1 , -1/2)

test value x = - 0.75

We now evaluate at x = - 0.75 to find its sign.

(positive)

c) interval (-1/2 , 4/5)

test value x = 0

We now evaluate at x = 0 to find its sign.

(negative)

d) interval (4/5 , 4)

test value x = 1

We now evaluate at x = 1 to find its sign.

(positive)

d) interval (4 , + )

test value x = 5

We now evaluate at x = 5 to find its sign.

(positive)



Let us now put all the above results in a table

-∞

-1   -1/2   4/5   4

+

Undefined + 0 - 0 + Undefined +

  • 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.



    Below is shown the graph to the function y = 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 5

    More rational inequalities with detailed examples.

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