How to Solve Second Degree Inequalities

2024 Author: Samantha Chapman | [email protected]. Last modified: 2024-01-15 13:47

The classic form of a second degree inequality is: ax ² + bx + c 0). Solving the inequality means finding the values of the unknown x for which the inequality is true; these values constitute the set of solutions, expressed in the form of an interval. There are 3 main methods: the straight line and verification point method, the algebraic method (most common) and the graphical one.

Steps

Part 1 of 3: Four Steps to Solving Second Degree Inequalities

Step 1. Step 1

Transform the inequality into a trinomial function f (x) on the left and leave 0 on the right.

Example. The inequality: x (6 x + 1) <15 is transformed into a trinomial as follows: f (x) = 6 x ² + x - 15 <0.

Step 2. Step 2

Solve the second degree equation to get the real roots. In general, a second degree equation can have zero, one or two real roots. You can:

• use the solution formula of second degree equations, or quadratic formula (it always works)
• factorize (if the roots are rational)
• complete the square (always works)
• draw the graph (for approximation)
• proceed by trial and error (shortcut for factoring).

Step 3. Step 3

Solve the second degree inequality, based on the values of the two real roots.

• You can choose one of the following methods:

  • Method 1: Use the line and verification point method. The 2 real roots are marked on the number line and divide it into a segment and two rays. Always use the origin O as a verification point. Substitute x = 0 into the given quadratic inequality. If it is true, the origin is placed on the correct segment (or radius).
  • Note. With this method, you could use a double line, or even a triple line, to solve systems of 2 or 3 quadratic inequalities into one variable.

  • Method 2. Use the theorem on the sign of f (x), if you have chosen the algebraic method. Once the development of the theorem has been studied, it is applied to solve various second degree inequalities.

    • Theorem on the sign of f (x):

      • Between 2 real roots, f (x) has the opposite sign to a; which means that:
      • Between 2 real roots, f (x) is positive if a is negative.
      • Between 2 real roots, f (x) is negative if a is positive.
      • You can understand the theorem by looking at the intersections between the parabola, the graph of the function f (x), and the axes of x. If a is positive, the parable faces upwards. Between the two points of intersection with x, a part of the parabola is under the axes of x, which means that f (x) is negative in this interval (of opposite sign to a).
      • This method may be faster than that of the number line because it does not require you to draw it every time. Furthermore, it helps to set up a table of signs for solving second degree systems of inequalities through the algebraic approach.

    

    Step 4. Step 4

    Express the solution (or set of solutions) in the form of intervals.

    • Examples of ranges:
    • (a, b), open interval, the 2 extremes a and b are not included
    • [a, b], closed interval, the 2 extremes are included

    • (-infinite, b], half closed interval, extreme b is included.

      Note 1. If the second degree inequality has no real roots, (discriminant Delta <0), f (x) is always positive (or always negative) depending on the sign of a, which means that the set of solutions will be o empty or will constitute the entire line of real numbers. If, on the other hand, the discriminant Delta = 0 (and therefore the inequality has a double root), the solutions can be: empty set, single point, set of real numbers {R} minus a point or the whole set of real numbers

    • Example: solve f (x) = 15x ^ 2 - 8x + 7> 0.
    • Solution. The discriminant Delta = b ^ 2 - 4ac = 64 - 420 0) regardless of the values of x. The inequality is always true.
    • Example: solve f (x) = -4x ^ 2 - 9x - 7> 0.

    • Solution. The discriminant Delta = 81 - 112 <0. There are no real roots. Since a is negative, f (x) is always negative, regardless of the values of x. The inequality is always not true.

      Note 2. When the inequality also includes a sign of equality (=) (greater and equal to or less than and equal to), use closed intervals such as [-4, 10] to indicate that the two extremes are included in the set of solutions. If the inequality is strictly major or strictly minor, use open intervals such as (-4, 10) since the extremes are not included

    Part 2 of 3: Example 1

    

    Step 1. Solve:

    15> 6 x ² + 43 x.

    

    Step 2. Transform the inequality into a trinomial

    f (x) = -6 x ² - 43 x + 15> 0.

    

    Step 3. Solve f (x) = 0 by trial and error

    • The rule of signs says that 2 roots have opposite signs if the constant term and the coefficient of x ² they have opposite signs.
    • Write down sets of probable solutions: {-3/2, 5/3}, {-1/2, 15/3}, {-1/3, 15/2}. The product of the numerators is the constant term (15) and the product of the denominators is the coefficient of the x term ²: 6 (always positive denominators).
    • Calculate the cross sum of each set of roots, possible solutions, by adding the first numerator multiplied by the second denominator to the first denominator multiplied by the second numerator. In this example, the cross sums are (-3) * (3) + (2) * (5) = 1, (-1) * (3) + (2) * (15) = 27 and (-1) * (2) + (3) * (15) = 43. Since the cross sum of the solution roots must be equal to - b * sign (a) where b is the coefficient of x and a is the coefficient of x ², we will choose the third together but we will have to exclude both solutions. The 2 real roots are: {1/3, -15/2}

    

    Step 4. Use the theorem to solve the inequality

    Between the 2 royal roots

    • f (x) is positive, with the opposite sign to a = -6. Outside this range, f (x) is negative. Since the original inequality had a strict inequality, it uses the open interval to exclude the extremes where f (x) = 0.

      The set of solutions is the interval (-15/2, 1/3)

    Part 3 of 3: Example 2

    

    Step 1. Solve:

    x (6x + 1) <15.

    

    Step 2. Transform the inequality into:

    f (x) = 6x ^ 2 + x - 15 <0.

    

    Step 3. The two roots have opposite signs

    

    Step 4. Write the probable root sets:

    (-3/2, 5/3) (-3/3, 5/2).

    • The diagonal sum of the first set is 10 - 9 = 1 = b.
    • The 2 real roots are 3/2 and -5/3.

    

    Step 5. Choose the number line method to solve the inequality

    

    Step 6. Choose the origin O as the verification point

    Substitute x = 0 into the inequality. It turns out: - 15 <0. It's true! The origin is therefore located on the true segment and the set of solutions is the interval (-5/3, 3/2).

    

    Step 7. Method 3

    Solve the second degree inequalities by drawing the graph.

    • The concept of the graphic method is simple. When the parabola, graph of the function f (x), is above the axes (or the axis) of x, the trinomial is positive, and vice versa, when it is below, it is negative. To solve second degree inequalities you will not need to graph the parabola with precision. Based on the 2 real roots, you can even just make a rough sketch of them. Just make sure that the dish is facing correctly downwards or upwards.
    • With this method you can solve systems of 2 or 3 quadratic inequalities, drawing the graph of 2 or 3 parabolas on the same coordinate system.

    Advice

    • During the checks or exams, the time available is always limited and you will have to find the set of solutions as quickly as possible. Always choose as the verification point the origin x = 0, (unless 0 is a root), as there is no time to verify with other points, nor to factor the second degree equation, recompose the 2 real roots in binomials, or discuss the signs of the two binomials.
    • Note. If the test, or exam, is structured with multiple choice answers and does not require an explanation of the method used, it is advisable to solve the quadratic inequality with the algebraic method because it is faster and does not require the drawing of the line.