Table of Contents

Last modified on May 13th, 2024

If we replace a quadratic equation’s equality sign (=) in the standard form ax^{2} + bx + c = 0 with an inequality sign, it becomes a quadratic inequality.

Here are a few examples of quadratic inequalities:

5x^{2} – 11x + 6 > 0

x^{2} + 5x – 6 < 0

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

- ax
^{2}+ bx + c > 0 - ax
^{2}+ bx + c < 0 - ax
^{2}+ bx + c â‰¥ 0 - ax
^{2}+ 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 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 5x^{2} – 11x + 6 > 0.

**Step 1. Writing in Standard Form**

5x^{2} – 11x + 6 > 0, which is in its standard form (ax^{2} + 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) = 5x^{2} – 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

5x^{2} – 11x + 6 = 0

â‡’ 5x^{2} – 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) = 5x^{2} – 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

Let us solve the inequality x^{2} + 5x – 6 < 0

**Step 1.** **Writing in Standard Form**

x^{2} + 5x – 6 < 0, which is in its standard form of ax^{2} + bx + c < 0.

**Step 2.** **Determining the Critical Points**

Factoring the inequality, we get

x^{2} + 5x – 6 < 0

â‡’ x^{2} + 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.

**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 + 3w^{2}

Since the area cannot exceed 120 m^{2}

Thus, 18w + 3w^{2} â‰¤ 120

â‡’ 3w^{2} + 18w – 120 â‰¤ 0

â‡’ w^{2} + 6w – 40 â‰¤ 0

â‡’ w^{2} + 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 > x ^{2} + 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

x^{2} + 3x + 2

= x^{2} + 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 May 13th, 2024