Quadratic inequalities ☰
A quadratic inequality is a type of inequality in which a quadratic function appears. Quadratic inequalities have the form:
\( ax^2 + bx + c,,,,op,,,,0 \), where \(a\), \(b\) and \(c\) are real numbers and \(op\) is one of the inequality symbols \( < \), \( \le \), \(> \) or \( \ge \).
The solution to a quadratic inequality is the set of all values of \(x\) that satisfy the inequality. To find the solution set, we can use the following steps:
- Solve the corresponding quadratic equation \( ax^2 + bx + c = 0 \) to find the roots of the quadratic function.
- Plot the graph of the quadratic function on a coordinate plane.
- Determine the region(s) of the graph that are above or below the \(x\)-axis, depending on the sign of \(a\).
- Shade the appropriate region(s) of the graph, depending on the inequality symbol.
- Write the solution set using interval notation or set notation.
Let's look at some examples:
Example 1: Solve the inequality \(x^2 - 4x + 3 \ge 0 \).
Solution: First, we find the roots of the corresponding quadratic equation \(x^2 - 4x + 3 = 0 \) by factoring or using the quadratic formula:
\( x^2 - 4x + 3 = (x-1)(x-3)=0 \).
Thus, the roots are \( x=1 \) and \(x=3 \).
Next, we plot the graph of the quadratic function \( f(x) = x^2 - 4x + 3 \).
Graph
We can see that the graph of the function is above the \(x\)-axis between the roots \(x=1\) and \(x=3\).
Therefore, the solution to the inequality \( x^2 - 4x + 3 \ge 0 \) is: \( x \in (-\infty, 1] \cup [3, \infty) \).
Example 2: An example of solving a quadratic inequality by the interval method.
Let's consider the inequality: \( 2x^2 - 5x - 3 < 0 \).
To solve this inequality, we first find the roots of the quadratic equation \( 2x^2–5x–3=0 \)
\( x=- \frac{1}{2} \) and \( x = 3 \). We can find these roots by factoring the quadratic or by using the quadratic formula.
Next, we use the roots to divide the number line into three intervals:
\( (-\infty ; -\frac{1}{2} ) \), \( (-\frac{1}{2}; 3) \) and \( (3; \infty)\).
We then test a value from each interval in the inequality to determine whether it is true or false in that interval.
For the interval \( (-\infty ; -\frac{1}{2}) \) we can choose \( x= -1 \) as a test value.
Substituting \(x=-1\) into the inequality, we get: \( 2(-1)^2 - 5(-1) - 3 < 0 \), which simplifies to: \( 4 < 0 \).
This is false, so the inequality does not hold in the interval \( (-\infty ; -\frac{1}{2}) \) .
For the interval \( (-\frac{1}{2} ; 3) \) we can choose \(x=0\) as a test value.
Substituting \(x=0\) into the inequality, we get: \( 2(0)^2 - 5(0) - 3 < 0 \), which simplifies to: \( -3 < 0 \).
This is true, so the inequality holds in the interval \( (-\frac{1}{2} ; 3) \).
For the interval \( (3; \infty) \) we can choose \(x=4\) as a test value.
Substituting \(x=4\) into the inequality, we get: \(2(4)^2 - 5(4) - 3 < 0 \), which simplifies to: \( 5 < 0 \) .
This is false, so the inequality does not hold in the interval \( (3;\infty) \).
Therefore, the solution to the inequality \( 2x^2–5x–3 < 0 \) is: \( x\ \in (- \frac{1}{2} ; 3) \)
This means that any value of \(x\) that is greater than \(- \frac{1}{2} \) and less than \(3\) makes the inequality true.
Solving rational inequalities by the interval method ☰
To solve rational inequalities by the interval method, follow these steps:
- Rewrite the inequality as a single rational expression, with zero on one side, and the other side in one fraction.
- Determine the critical values of the expression by finding where the numerator and denominator are equal to zero.
- Divide the number line into intervals separated by the critical values found in step 2.
- Determine the sign of the expression in each interval, and identify the intervals where the expression is positive or negative.
- Write the solution in interval notation, using the intervals where the expression is positive or negative, depending on the direction of the inequality.
For example, let's solve the inequality \(\frac{2x-5}{x+1} > 1 \) by the interval method.
1. Rewrite the inequality as a single rational expression:
\(\frac{2x-5}{x+1} -1 > 0 \)
Simplify the left-hand side and combine like terms:
\( \frac{2x-5-(x+1)}{x+1} > 0 \)
\( \frac{x-6}{x+1} \)
2. Determine the critical values by setting the numerator and denominator equal to zero:
\( x-6=0 \rightarrow x=6 \).
\( x+1=0 \rightarrow x=-1 \).
3. Divide the number line into intervals:
\( (-\infty ; -1) \), \( (-1;6) \), \( (6;\infty ) \).
4. Determine the sign of the expression in each interval by testing a point in each interval:
For \( x=-2 \), \( \frac{-2-6}{-2+1} = 8 > 0 \) , so the expression is positive in \( (-\infty ; -1) \).
For \( x = 0 \), \( \frac{0-6}{0+1} = -6 < 0 \) , so the expression is negative in \( (-1 ; 6 ) \).
For \( x = 7 \), \( \frac{7-6}{7+1} = \frac{1}{8} > 0 \) , so the expression is positive in \( (6; \infty ) \).
5. Write the solution in interval notation:
The inequality is true where the expression is positive, so the solution is \( x \in (- \infty ; -1 ) \cup (6; \infty ) \).
Irrational inequalities ☰
An irrational inequality is an inequality involving one or more irrational expressions, such as square roots or cube roots. The process for solving an irrational inequality is different from that of solving a regular inequality, because squaring or raising to a power may introduce extraneous solutions. To solve an irrational inequality, you need to isolate the irrational expression on one side of the inequality, and then square or raise both sides of the inequality to eliminate the radical. However, when you do this, you may introduce solutions that are not valid for the original inequality, because squaring or raising to a power can change the sign of a number. Therefore, you need to check your solutions to make sure they are valid for the original inequality.
To solve irrational inequalities, follow these general steps:
1. Isolate the irrational expression on one side of the inequality.
2. Square both sides of the inequality (or raise both sides to a power that eliminates the radical).
3. Solve the resulting inequality.
4. Check the solutions to the original inequality, since squaring or raising to a power may introduce extraneous solutions.
For example, let's solve the inequality \( \sqrt{2x-1} > 3 \) .
1. Square both sides of the inequality:
\( (\sqrt{2x-1})^2 > 3^2 \rightarrow 2x - 1 > 9 \)
2. Solve the resulting inequality:
\( 2x > 10 \rightarrow x > 5 \) .
\( \small \sqrt{2x-1} > 3 \rightarrow 2x > 10 \rightarrow x > 5 \)
3. Check the solution to the original inequality:
Substitute \(x=6\) into the original inequality:
\( \sqrt{2(6)-1} > 3 \rightarrow \sqrt{11} > 3 \)
This is true, so \(x=6\) is a valid solution to the inequality.
Substitute \(x=4\) into the original inequality:
\( \sqrt{2(4)-1} > 3 \rightarrow \sqrt{7} > 3 \)
This is false, so \(x=4\) is not a valid solution to the inequality.
Therefore, the solution is \(x > 5\).
Note: Always note that the root expression cannot be a negative number.
