The standard form of a quadratic inequality in one variable is f(x) = ax^2 + bx + c > 0 (or < 0). If the inequality is written in another form, transform it into the standard form, with a trinomial in x on the left side and 0 on the right side.
Solving the inequality for x means finding the values of x that make the inequality true. These values of x constitute the solution set of the inequality. The solution set is expressed in terms of intervals. By convention, we use these following symbols, x1 and x2 being the 2 real roots of the equation f(x) = 0:
- (x1 , x2) for the open interval between x1 and x2. The end points x1 and x2 are not included in the solution set.
- [x1 , x2] for the closed interval between x1 and x2. The end points are included in the solution set.
Note. When the inequality has an additional equal sign (greater, less, or equal to), the 2 end points are automatically included in the solution set. The solution set may be a closed interval [x1 , x2]. Or, the solution set may be half closed intervals: (-infinity, x1] and [x2 , + infinity).
There are 3 common methods to solve a quadratic inequality:
1. Using the number-line
2. Using the algebraic method.
3. By graphing.
Approaches in solving quadratic inequalities.
There are 2 approaches in solving quadratic inequalities depending on the chosen method. The test point approach is convenient for the number-line or the graphing method. The second approach uses a theorem on the sign of a trinomial f(x), if the algebraic method is chosen.
Examples of solving quadratic inequalities.
Example 1. Solve: f(x) = 12x^2 + 5x - 72 < 0.
Solution. First, solve the equation f(x) = 0. I advise you to use the Diagonal Sum Method to solve it. Roots have opposite signs. Write down the all-options-line. After elimination of no-fitted options, the remainder probable roots-set is (-8/3 & 9/4) that give as 2 real roots: -8/3 and 9/4. Next, solve the inequality f(x) < 0.
Plot the two real roots -8/3 and 9/4 on the number line. Use the origin as test point. Substitute x = 0 into the inequality. You get: -72 < 0. It is true, then the origin is located on the true segment. The solution set is the open interval: (-8/3 , 9/4). The 2 end points are not included in the solution set.
Example 2. Solve f(x) = 45x^2 + 74x - 55 > 0.
Solution. Use the Diagonal Sum Method to solve f(x) = 0. It gives 2 real roots:
-11/5 and 5/9. Plot these 2 values on the number-line. Substitute x = 0 into the inequality f(x) > 0. You get: -55 > 0. It is not true!. Then, the origin is on the "wrong" segment. The solution set are the two rays. The solution set are: (-infinity , -11/5) or (5/9 , +infinity). The 2 end points are not included.
A double inequality is an inequality where there are two signs, as opposed to one.Ex: an inequality could be 3x < 15A double inequality could be 3x < 15 < x + 20If you'd want to solve that double inequality, you split it into to expressions:3x < 15 and x + 20 > 15Then just solve.x < 5 and x > -5-5 < x < 5
Some signs of inequality may be physical or emotional differences.
In mathematics, an inequality is a statement about the relative size or order of two values.
When a quantity is subtracted or added from both sides of an inequality, the true difference in value is varied thereby changing the direction of the inequality, but when rather than subtracted or added it is multiplied or divided, it preserves the true difference in value thereby facing the same direction as the initial inequality.
Not the same as, or different from.
A linear inequality is all of one side of a plane. A quadratic inequality is either the inside of a parabola or the outside.
A quadratic inequality in x is in the standard form of ax^2+bx+c(>or<)d. Ex. 3x^2+5x+1>4
No. This is not an inequality, because you need something > something_else, or less than or 'not equal' or 'greater than or equal', etc. Since it has an x cubed term, it is not a quadratic.
In a linear inequality the variable is only present raised to the first power (which is usually not explicitly shown). In a quadratic the square of the variable is present (or implied). The square can be implied in an inequality such as x + 1/x < 6 (x not 0) This is equivalent to x2 - 6x + 1 < 0
The answer depends on the nature of the inequality: whether it is linear, quadratic or has some other functional form.
Use the quadratic formula for the equality. Then, depending on the coefficient of x2 and the nature of the inequality [>, ≥, ≤, <], determine whether you need the open or closed intervals between the roots or beyond the roots.
Yes. Consider x2 ≥ 0
Yes - except in extreme cases. It can be the whole of the Real Numbers: eg x2 > -3 It can be a single point eg x2 ≤ 0 gives x = 0
It could be the solution to some quadratic inequalities: for example x2 + 2x - 3 > 0 whose solution is x < -3 or x > 1.
The main difference is that if you multiply both sides of an inequality by a negative number, you have to change the direction of the inequality sign - for example "greater than" would become "less than".
The quadratic formula is used to solve the quadratic equation. Many equations in which the variable is squared can be written as a quadratic equation, and then solved with the quadratic formula.
Substitute the values of the variables into the inequality. If the inequality is true then they are a solution, if not, they are not.Substitute the values of the variables into the inequality. If the inequality is true then they are a solution, if not, they are not.Substitute the values of the variables into the inequality. If the inequality is true then they are a solution, if not, they are not.Substitute the values of the variables into the inequality. If the inequality is true then they are a solution, if not, they are not.