Table of Contents
Last modified on April 25th, 2024
If we replace a quadratic equation’s equality sign (=) in the standard form ax2 + bx + c = 0 with an inequality sign, it becomes a quadratic inequality.
Here are a few examples of quadratic inequalities:
5x2 – 11x + 6 > 0
x2 + 5x – 6 < 0
Depending on the sign, the 4 standard forms of quadratic inequality are:
Here, like a quadratic equation, ‘a’ (≠0), ‘b,’ and ‘c’ are the constants, and ‘x’ is a variable.
Solving quadratic inequalities involves finding the value of the variable that satisfies the inequality. It can be done both graphically and algebraically.
While solving quadratic inequalities graphically, we find the intervals where the inequality is either above or below the x-axis, depending on the inequality sign.
To understand the concept, let us solve the inequality 5x2 – 11x + 6 > 0.
Step 1. Writing in Standard Form
5x2 – 11x + 6 > 0, which is in its standard form (ax2 + bx + c > 0).
Step 2. Graphing the Quadratic Function
Finding the Line of Symmetry
${x=-\dfrac{b}{2a}}$
⇒ ${x=-\dfrac{-11}{2\times 5}}$
⇒ ${x=\dfrac{11}{10}}$
Finding the Vertex of the Quadratic Function
f(x) = 5x2 – 11x + 6,
${f\left( \dfrac{11}{10}\right) =5\left( \dfrac{11}{10}\right) ^{2}-11\left( \dfrac{11}{10}\right) +6}$
⇒ ${f\left( \dfrac{11}{10}\right) =\dfrac{605}{100}-\dfrac{121}{10}+6}$
⇒ ${f\left( \dfrac{11}{10}\right) =\dfrac{-5}{100}}$
⇒ ${f\left( \dfrac{11}{10}\right) =\dfrac{-1}{20}}$
Thus, the vertex is ${\left( \dfrac{11}{10},\dfrac{-1}{20}\right)}$ ⇒ (1.1, -0.05)
Finding the x-intercepts
By putting f(x) = 0, we get
5x2 – 11x + 6 = 0
⇒ 5x2 – 5x – 6x + 6 = 0
⇒ (x – 1)(5x – 6) = 0
⇒ x = 1, ${\dfrac{6}{5}}$
⇒ x = 1, 1.2
Thus, x-intercepts are (1, 0) and (1.2, 0)
Now, graphing the quadratic function f(x) = 5x2 – 11x + 6, we get
Step 4. Determining the Solution Set
Since the inequality has a ‘>’ sign, the vertices are not included, and the solution set is (-∞, 1) ∪ (1.2, ∞). On graphing the solution, we get
Let us solve the inequality x2 + 5x – 6 < 0
Step 1. Writing in Standard Form
x2 + 5x – 6 < 0, which is in its standard form of ax2 + bx + c < 0.
Step 2. Determining the Critical Points
Factoring the inequality, we get
x2 + 5x – 6 < 0
⇒ x2 + 6x – x – 6 < 0
⇒ (x + 6)(x – 1) < 0
Now, finding the critical points (the solutions) to the related quadratic equation, we get
(x + 6)(x – 1) = 0
⇒ x = -6, 1
Step 3. Dividing the Number Line into Intervals with the Critical Points
Step 4. Determining the Sign of the Expression for Distinct Intervals
At x = -10 (< -6), (-10)2 + 5(-10) – 6 = 44 ≮ 0
At x = -1 (> -6 but < 1), (-1)2 + 5(-1) – 6 = -10 < 0
At x = 5 (> 1), (5)2 + 5(5) – 6 = 44 ≮ 0
Thus, at -6 < x < 1, the inequality holds.
Step 5. Representing the Solution Set on the Number Line
Here, the solution set of the quadratic inequality is (-6, 1), the interval notation.
The length of a rectangular box is 18 meters more than thrice its width. Find all possible rectangle widths that would result in an area of no more than 120 square meters and graph the solution on a number line.
Let ‘l’ and ‘w’ be the length and width of the rectangular box.
As we know, the length is 18 meters, more than thrice the width
Here, l = 18 + 3w
Area = l â‹… w
=Â w(18 + 3w)
= 18w + 3w2
Since the area cannot exceed 120 m2
Thus, 18w + 3w2 ≤ 120
⇒ 3w2 + 18w – 120 ≤ 0
⇒ w2 + 6w – 40 ≤ 0
⇒ w2 + 10w – 4w – 40 ≤ 0
⇒ (w + 10)(w – 4) ≤ 0
Now, finding the critical points (the solutions) to the related quadratic equation, we get
(w + 10)(w – 4) = 0
⇒ w = -10, 4
Here, the interval points are -10 and 4.
Now, at w = -11 (< -10), 3(-11)2 + 18(-11) = 363 – 198 = 165, which is not less than 120
At w = 0 (> -10 but < 4), 3(0)2 + 18(0) = 0, which is less than 120
At w = 5 (> 4), 3(5)2 + 18(5) = 165, which is not less than 120
Thus, at -10 < w < 4, the inequality holds.
Since the width cannot be negative, the only valid solution is 0 < w ≤ 4
Thus, all possible rectangle widths resulting in an area of no more than 120 square meters are w ≤ 4 meters, which means w = {1, 2, 3, 4}
Which quadratic inequality does the graph below represent?
a) y > x2 + 3x + 2
b) y < x2 + 3x + 2
c) y > x2 – 3x – 2
d) y < x2 – 3x – 2
As we know, the y-intercept is 2, not -2
The options c) and d) are eliminated.
Now, on factoring the quadratic expressions,Â
x2 + 3x + 2
= x2 + 2x + x + 2
= (x + 2)(x + 1)
The x-intercepts are (-1, 0) and (-2, 0)
Since the shaded part is above the curve.
Thus, a) is the correct quadratic inequality, which represents the given graph.
Last modified on April 25th, 2024