Quadratic Inequalities - Definition, Expression, Graphs, Solved Examples

Inequalities are mathematical expressions showing the relationship between two values, indicating that one value is greater than, less than, or not equal to another. Understanding inequalities is crucial for solving various mathematical problems, from basic arithmetic to advanced calculus. Quadratic inequalities are specially used for a parabola that is two-degree curves.

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In this article, we will cover the concept of the sign of quadratic expression. This concept falls under the broader category of complex numbers and quadratic equations, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

Method to solve quadratic inequalities

The wavy curve method is used to solve the inequality of the type. Inequalities can be solved by drawing their graph and before that converting them into standard form with zero at right hand side. Then drawing the graph accordingly with area above x axis bankor below

$$
\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}>0 \text { or }(<, \leq, \geq)
$$

We use the following steps in the wavy curve method to solve a question (start by getting 0 on one side of inequality)
1: Factorize the numerator and denominator into linear factors.
2: Make coefficients of $x$ positive in all linear factors.
3: Equate each linear factor to zero and find the values of $x$ in each case. The values are called critical points. Do not include the linear factors with even power while finding critical points.
4: Identify distinct critical points on the real number line. The " $n$ " numbers of distinct critical points divide real number lines in ( $n+1$ ) sub-intervals.
5: The sign of rational function in the rightmost interval is positive. Alternate sign in adjoining intervals on the left.
6. Check for each critical point and point from even powered linear factors, if these are to be included in the answer or not.

For Example: using the wavy curve method to find the interval of $x$ for the inequality given :

$$
\frac{x}{x-1} \geq 0
$$

Steps
- 0 on the right-hand side (already there)
- all linear factors (already present)
- Critical points are: $x=0,1$

The critical points are marked on the real number line. Starting with a positive sign in the rightmost interval, we denote signs of adjacent intervals by alternating signs.

Hence, $x \in(-\infty, 0] \cup(1, \infty)$
Note that 1 cannot be taken in the answer as at $x=1$, the denominator becomes 0 , and hence expression is not defined.

Summary

Inequalities are a fundamental part of mathematics, providing a way to describe and solve problems involving ranges and constraints. Mastery of inequalities is essential for progressing in algebra, calculus, and applied mathematics, offering valuable tools for both theoretical and practical problem-solving.

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Solved Examples Based On the Inequalities:
Example 1: The least value of $|z|_{\text {where } \mathbf{z}}$ is a complex number that satisfies the inequality $\exp \left(\frac{(|z|+3)(|z|-1)}{|| z|-1|} \log _e 2\right) \geqslant \log _{\sqrt{2}}|5 \sqrt{7}+9 i|, i=\sqrt{-1}$,is equal to:
Solution:

$$
\begin{aligned}
& \exp \left(\frac{(|z|+3)(|z|-1)}{|| z|+1|} \ln 2\right) \geq \log _{\sqrt{2}}|5 \sqrt{7}+9 i| \\
& \Rightarrow 2^{\frac{(|z|+3)(|z|-1)}{(\mid z+1)} \geq \log _{\sqrt{2}}(16)} \\
& \Rightarrow 2^{\frac{(|z|+3)(|z|-1)}{(z \mid+1)} \geq 2^3} \\
& \Rightarrow \frac{(|z|+3)(|z|-1)}{(z \mid+1)} \geq 3 \\
& \Rightarrow(|z|+3)(|z|-1) \geq 3(|z|+1) \\
& |z|^2+2|z|-3 \geq 3|z|+3
\end{aligned}
$$

$$
\begin{aligned}
& \Rightarrow|z|^2+|z|-6 \geq 0 \\
& \Rightarrow(|z|-3)(|z|+2) \geq 0 \Rightarrow|z|-3 \geq 0 \\
& \Rightarrow|z| \geq 3 \quad \Rightarrow|z|_{\min }=3
\end{aligned}
$$

Hence, the answer is 3.
Example 2: The integer ' k ', for which the inequality $x^2-2(3 k-1) x+8 k^2-7>0$ is valid for every real value of x , is
Solution:
The condition for a quadratic expression to be always positive is
a (coefficient of $x^2$ ) $>0$ and $D<0$
Now,
Coefficient of $x^2=1(>0)$
So, we have to apply only $\mathrm{D}<0$

$$
\begin{aligned}
& \Rightarrow 4(3 \mathrm{~K}-1)^2-4 \times 1 \times\left(8 \mathrm{~K}^2-7\right)<0 \\
& \Rightarrow 9 \mathrm{~K}^2-6 \mathrm{~K}+1-8 \mathrm{~K}^2+7<0 \\
& \Rightarrow \mathrm{K}^2-6 \mathrm{~K}+8<0 \\
& \Rightarrow(\mathrm{K}-4)(\mathrm{K}-2)<0 \\
& \Rightarrow \mathrm{K} \in(2,4)
\end{aligned}
$$

So, only one integer in this interval: $\mathrm{K}=3$.
Hence, the answer is 3 .
Example 3: The solution of $(x+1)(x-9)<0$ is
1) $[-\infty,-1] \cup(9-\infty)$
2) $(-\infty,-9) \cup(1, \infty)$
3) $[-1,9]$
4) $(-1,9)$

Solution:

$$
(x+1)(x-9)<0
$$

As the inequation is in standard form, we can make the sign scheme.
Roots: $-1,9$

So, the solution is (-1,9)

Example 4: Solution of $x(x-4)(x+5)>0$
1) $(-5,0) \cup(4, \infty)$
2) $(-\infty,-5) \cup(4, \infty)$
3) $(-5, \infty)$
4) None of these

Solution:
Already in standard form
Roots $x=0,4,-5$
Sign scheme

So, $(-5,0) \cup(4, \infty)$
Example 5: The solution of $\frac{x^3(x-4)(x+3)^2}{(x+1)} \geqslant 0$ is
1) $(-1,0] \cup[4, \infty)$
2) $(-1,0] \cup[4, \infty) \cup-3$
3) $(-1,0) \cup(4, \infty)$
4) $(-\infty,-3] \cup(-1,0] \cup[4, \infty)$

Solution:
Standard form
Roots: $x=0,4,-3$
Roots of denominator: -1
Sign scheme

Positive for $(-1,0) \cup(4, \infty)$
Zero for $0,4,-3$
So, the solution is $(-1,0] \cup[4, \infty) \cup-3$
Frequently Asked Questions(FAQ)-
1. What is inequalities?

Ans: Inequalities are the relationship between two expressions which are not equal to one another.
2. Solve $-5(x-1) \leq 10(2 x-3)$

Ans:

$$
\begin{aligned}
& -5(x-1) \leqslant 010(2 x-3) \\
\Rightarrow \quad & (x-1) \geqslant \frac{10}{-5}(2 x-3) \\
\Rightarrow & x-1 \geqslant-2(2 x-3) \\
\Rightarrow & x-1 \geqslant-4 x+6 \\
\Rightarrow & x+4 x \geqslant 6+1 \\
\Rightarrow & 5 x \geqslant 7 \\
\Rightarrow & x \geqslant \frac{7}{5} \\
& \frac{-1}{2}<x \leq 3, \text { then } \frac{1}{x}
\end{aligned}
$$

Ans:

$$
\begin{aligned}
& -\frac{1}{2}<x \leqslant 3\left(-\frac{1}{2}<0,3>0\right) \\
& \Rightarrow-\frac{1}{2}<x<0^{-} \text {or } 0^{+}<x \leqslant 3 \\
& \Rightarrow-2>\frac{1}{x}>-\infty \text { or } \infty>\frac{1}{x} \geqslant \frac{1}{3} \\
& \Rightarrow \frac{1}{x} \in(-\infty,-2) \cup\left[\frac{1}{3}, \infty\right)
\end{aligned}
$$

4. What is the solution of the expression $x^2<9$ ?

Ans: $x^2<9 \Rightarrow x^2-9<0 \Rightarrow(x-3)(x+3)<0$

$$
x \in(-3,3)
$$

5. The equation $e^{4 \mathrm{x}}+8 \mathrm{e}^{3 \mathrm{x}}+13 \mathrm{e}^{2 \mathrm{x}}-8 \mathrm{e}^{\mathrm{x}}+1=0, \mathrm{x} \in \mathbb{R}_{\text {has }}$ :

Ans: 1

$$
\mathrm{e}^{4 x}+8 \mathrm{e}^{3 x}+13 \mathrm{e}^{2 x}+13 \mathrm{e}^{2 x}-8 \mathrm{e}^{\mathrm{x}}+1=0, x \in R
$$

Let $\mathrm{e}^{\mathrm{x}}=\mathrm{t}>0 \& \mathrm{x}=\ln t$

$$
t^4+8 t^3+13 t^2-8 t+1=0
$$

Dividing by $\mathrm{t}^2$,

$\begin{aligned} & t^2+8 t+13-\frac{8}{t}+\frac{1}{t^2}=0 \\ & t^2+\frac{1}{t^2}+8\left(t-\frac{1}{t}\right)+13=0 \\ & \quad \mathrm{t}-\frac{1}{\mathrm{t}}=\mathrm{u} \Rightarrow \mathrm{t}^2+\frac{1}{t^2}-2 \mathrm{u}^2 \\ & \text { Let } \\ & \Rightarrow \mathrm{t}^2+\frac{1}{\mathrm{t}^2}=\mathrm{u}^2+2 \\ & \mathrm{u}^2+2+8 \mathrm{u}+13=0 \\ & (\mathrm{u}+3)(\mathrm{u}+5)=0 \\ & \mathrm{u}=-3 \& \mathrm{u}=-5 \\ & \mathrm{t}-\frac{1}{\mathrm{t}}=-3 \\ & \mathrm{t}^2+3 \mathrm{t}-1=0 \quad \mathrm{t}-\frac{1}{\mathrm{t}}=-5 \\ & \mathrm{t}^2+5 \mathrm{t}-1=0\end{aligned}$

$$
\begin{aligned}
& 0<\alpha_1<1 \quad 0<\alpha_2<1 \\
& \Rightarrow \mathrm{x}_1=\ln \alpha_1<0 \quad \Rightarrow \mathrm{x}_2=\ln \alpha_2<0
\end{aligned}
$$