### #ezw_tco-2 .ez-toc-title{ font-size: 120%; ; ; } #ezw_tco-2 .ez-toc-widget-container ul.ez-toc-list li.active{ background-color: #ededed; } chapter outline

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

## Standard Forms

Depending on the sign, the 4 standard forms of quadratic inequality are:

• ax2 + bx + c > 0
• ax2 + bx + c < 0
• ax2 + bx + c â‰¥ 0
• ax2 + bx + c â‰¤ 0

Here, like a quadratic equation, â€˜aâ€™ (â‰  0), â€˜b,â€™ and â€˜câ€™ are the constants, and â€˜xâ€™ is a variable.

## Solving

Solving quadratic inequalities involves finding the value of the variable that satisfies the inequality. It can be done both graphically and algebraically.

### Graphically

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

${x=-\dfrac{b}{2a}}$

â‡’ ${x=-\dfrac{-11}{2\times 5}}$

â‡’ ${x=\dfrac{11}{10}}$

Finding the Vertex of the Quadratic Function

f(x) = 5x2 – 11x + 6,

${f\left( \dfrac{11}{10}\right) =5\left( \dfrac{11}{10}\right) ^{2}-11\left( \dfrac{11}{10}\right) +6}$

â‡’ ${f\left( \dfrac{11}{10}\right) =\dfrac{605}{100}-\dfrac{121}{10}+6}$

â‡’ ${f\left( \dfrac{11}{10}\right) =\dfrac{-5}{100}}$

â‡’ ${f\left( \dfrac{11}{10}\right) =\dfrac{-1}{20}}$

Thus, the vertex is ${\left( \dfrac{11}{10},\dfrac{-1}{20}\right)}$ â‡’ (1.1, -0.05)

Finding the x-intercepts

By putting f(x) = 0, we get

5x2 – 11x + 6 = 0

â‡’ 5x2 – 5x – 6x + 6 = 0

â‡’ (x – 1)(5x – 6) = 0

â‡’ x = 1, ${\dfrac{6}{5}}$

â‡’ 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 â€˜>â€™ sign, the vertices are not included, and the solution set is (-âˆž, 1) âˆª (1.2, âˆž). On graphing the solution, we get

### Algebraically

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

â‡’ x2 + 6x – x – 6 < 0

â‡’ (x + 6)(x – 1) < 0

Now, finding the critical points (the solutions) to the related quadratic equation, we get

(x + 6)(x – 1) = 0

â‡’ 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 â‰® 0

At x = -1 (> -6 but < 1), (-1)2 + 5(-1) – 6 = -10 < 0

At x = 5 (> 1), (5)2 + 5(5) – 6 = 44 â‰® 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.

## Solved Examples

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.

Solution:

Let â€˜lâ€™ and â€˜wâ€™ 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 â‰¤ 120
â‡’ 3w2 + 18w – 120 â‰¤ 0
â‡’ w2 + 6w – 40 â‰¤ 0
â‡’ w2 + 10w – 4w – 40 â‰¤ 0
â‡’ (w + 10)(w – 4) â‰¤ 0
Now, finding the critical points (the solutions) to the related quadratic equation, we get
(w + 10)(w – 4) = 0
â‡’ 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 â‰¤ 4
Thus, all possible rectangle widths resulting in an area of no more than 120 square meters are w â‰¤ 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

Solution:

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.