Irrational equations and Inequalities: Problems with Solutions

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. Irrational equations are help in solving fractional power equations.

In this article, we will cover the concepts of irrational inequalities and equations. This concept falls under the broader category of complex numbers., 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. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of three questions have been asked on this concept, including one in 2016, one in 2018, and one in 2019.

Inequalities

Inequalities are the relationship between two expressions that are not equal to one another. Symbols denoting the inequalities are <, >, ≤, ≥, and ≠.

• $x<4$, "is read as $x$ less than $4^{\prime \prime}, x \leq 4$, is read as $x$ less than or equal to $4^{\prime \prime}$.
• Similarly $x>4$, "is read as $x$ greater than $4^{\circ}$ and $x \geq 4$, "is read as $x$ greater than or equal to 4 ".

The process of solving inequalities is the same as of equality but instead of equality symbol inequality symbol is used throughout the process.

Types of Inequalities

• Linear Inequalities: Involve linear expressions.
  • Example: $2 x+3 \leq 7$
• Quadratic Inequalities: Involve quadratic expressions.
  • Example: $x^2-4 x+3 \geq 0$
• Polynomial Inequalities: Involve polynomials of degree greater than two.
  • Example: $x^3-2 x^2+x-5<0$
• Rational Inequalities: Involve ratios of polynomials.
  • Example: $\frac{x+1}{x-3} \geq 2$
• Absolute Value Inequalities: Involve absolute value expressions.
  • Example: $|x-2| \leq 5$

A few rules that are different from equality rules

• - If we multiply or divide both sides of the inequality by a negative number, then we reverse the inequality (reversing inequality means > gets converted to < and vice versa, and $\geq$ gets converted to $\leq$ and vice versa) (eg $4>3$ means $-4<-3$ )
• If we cross multiply a negative quantity in an inequality, then we reverse the inequality (eg $3>-2$ means $-3 / 2<1$ )
• If we cancel the minus sign from both sides of an inequality, then we reverse the inequality. (eg $-3>-4$ means $3<4$ )
• As we usually do not know the sign of a variable term like $x,(x-2)$, etc, so we do not cross-multiply them, as we cannot decide if we have to reverse the sign of inequality or not.

We get a range of solutions while solving inequality which satisfies the inequality,

for e.g. a $>3$ gives us a range of solutions, means a ? $(3, \infty)$
Graphically inequalities can be shown as a region belonging to one side of the line or between lines, for example, inequality $-3<x \leq 5$ can be represented as below, a region belonging to $-3$ and $5$ are the region of possible $x$ including $45$ and excluding $-3$ .

Irrational equation

An irrational equation is an equation where the variable is inside the radical or the variable is a base of power with fractional exponents.

For example,

$\sqrt{2 x-3}=4$

To solve any irrational equation, eliminate the radical by raising the term in the other side to a power. For the equation given in the form of $\sqrt[n]{f(x)}=g(x)$, where $n \in N$ and $f(x), g(x)$ are polynomial function, solve $f(x)=(g(x))^n$,
and check the values of $x$ obtained by putting it in original equation

For example,

given equation is $\sqrt{x^2-4 x+4}=x+1$
then, $x^2-4 x+4=(x+1)^2$
$\Rightarrow \mathrm{x}^2-4 \mathrm{x}+4=\mathrm{x}^2+2 \mathrm{x}+1$
$\Rightarrow 6 \mathrm{x}=3 \Rightarrow \mathrm{x}=\frac{1}{2}$
$\mathrm{x}=\frac{1}{2}$ satisfies the original equation
so $\mathrm{x}=1 / 2$ is the answer

Irrational Inequalities:

Irrational inequalities are inequalities with irrational equations. For example, $\sqrt{x^2+7} = 4$

If $n$ is oddd

To solve inequations of the form $(f(x))^{1 / n}>g(x)$ or $(f(x))^{1 / n}<g(x)$, or $(f(x))^{1 / n}>(g(x))^{1 / n}$, raise both sides to the power $n$, and solve to get the answer.

If $n$ is even
1. To solve inequations of the form $(f(x))^{1 / n}>g(x)$,
a. LHS should be defined, so solve $f(x) \geq 0$
b. Now if $\mathrm{g}(\mathrm{x})<0$, then LHS will be greater than RHS for all such values
c. If $g(x) \geq 0$, then solve $f(x)>(g(x))^n$

In the end take intersection of a with (b union c)
2. To solve inequations of the form $(f(x))^{1 / n}<g(x)$,
a. LHS should be defined, so solve $f(x) \geq 0$
b. Now if $\mathrm{g}(\mathrm{x})<0$, then LHS will be not be lesser than RHS for all such values
c. If $g(x) \geq 0$, then solve $f(x)<(g(x))^n$

In the end take intersection of $a$ and $c$

For example,

$\begin{aligned} 3+\sqrt{3 x+1}=x & \Rightarrow \sqrt{3 x+1}=x-3 \\ & \Rightarrow 3 x+1=(x-3)^2 \\ & \Rightarrow 3 x+1=x^2-6 x+9 \\ & \Rightarrow 0=x^2-9 x+8 \\ & \Rightarrow x=8, x=1\end{aligned}$

Solved Examples Based On Irrational Inequalities and Equations:

Example 1: Solve the inequality $\sqrt{x+14}<x+2$

Solution:

$\sqrt{x+14}<x+2$

1. For LHS to be defined, $x+14 \geq 0$ means $x \geq-14$
2. When $x+2 \leqslant 0$, then RHS cannot be greater than LHS, so no answer from this case
3. When $x+2 \geq 0$, means when $x \geq-2$,

In this case, we can square both sides

$
\begin{aligned}
& x+14<(x+2)^2 \\
& x+14<x^2+4 x+4 \\
& x^2+3 x-10>0 \\
& (x+5)(x-2)>0 \\
& x<-5 \text { or } x>2
\end{aligned}
$

Taking intersection with $x \geq-2$, which equals $x>2$
Now, the answer is the intersection of (1) and (3), which is $x>2$

Example 2: Which of the options is correct for the inequality $\sqrt{-x^2+4 x-3}>6-2 x ?$

1) $(1,3)$
2) $\left(\frac{13}{5}, 3\right)$
3) $(3, \infty)$
4) $(-\infty, 3)$

Solution

1. LHS should be defined, so

$-x^2+4 x-3 \geq 0$

$\begin{aligned} & x^2-4 x+3 \leq 0 \\ & (x-1)(x-3) \leq 0 \\ & 1 \leq x \leq 3\end{aligned}$

2. When RHS < 0, then all these values will satisfy the inequation

$6-2 x<0$

$\Rightarrow x>2$

3. When RHS 0 (when $x \leq 3$ ), then we can square the inequation

$
\begin{aligned}
& -x^2+4 x-3>(6-2 x)^2 \\
& -x^2+4 x-3>36+4 x^2-24 x \\
& 5 x^2-28 x+39<0 \\
& (x-3)(5 x-13)<0 \\
& 13 / 5<x<3
\end{aligned}
$

Taking the intersection of this result with $x<3$, we get the interval of x , i.e

$13 / 5<x<3$

Hence, the answer is the option 2.

Example 3: If x is a solution of the equation , $\sqrt{2 x+1}-\sqrt{2 x-1}=1,\left(x \geqslant \frac{1}{2}\right)$, then $\sqrt{4 x^2-1}$ is equal to :

Solution:

$
\begin{aligned}
& \sqrt{2 x+1}-\sqrt{2 x-1}=1 \\
& \Rightarrow \sqrt{2 x+1}=\sqrt{2 x-1}+1
\end{aligned}
$

square both side

$
\begin{aligned}
& \Rightarrow 2 x+1=2 x-1+1+2 \sqrt{2 x-1} \\
& \Rightarrow 1=2 \sqrt{2 x-1}
\end{aligned}
$

square both side

$
\begin{aligned}
& \Rightarrow 2 x-1=\frac{1}{4} \\
& \Rightarrow x=\frac{5}{8}
\end{aligned}
$

Now $\sqrt{4 x^2-1}$ at $x=5 / 8 \Rightarrow \sqrt{4 \times \frac{25}{64}-1}=3 / 4$

Example 4: Let $S={ }^{-1}$ R: $x>0$ 0 and $2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0$ Then S:

1) contains exactly four elements

2) is an empty set.

3) contains exactly one element

4) contains exactly two elements

Solution:

As we learned in

Roots of Quadratic Equation -

$\alpha=\frac{-b+\sqrt{b^2-4 a c}}{2 a}$

$\beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a}$

- wherein

$a x^2+b x+c=0$

is the equation

$a, b, c \in R, \quad a \neq 0$

Case 1

$\sqrt{x} \geq 3 \Rightarrow x \geq 9$

$2(t-3)+t(t-6)+6=0$

$t^2-4 t=0$

$\Rightarrow t=0, t=4$

$\sqrt{x}=0, \sqrt{x}=4$

$x=0, x=16$

we take $x=16$ $x \geq 9$

case 2

$0<\sqrt{x}<3 \Rightarrow 0<x<9$

$-2 t+6+t^2-6 t+6=0$

$t^2-8 t+12=0$

$\Rightarrow t=2, t=6$

$\Rightarrow x=4, x=36$

Thus $x=4$ : $x<9$

So there are two elements

Hence, the answer is the option 4.

Example 5: Solve $x-3 \sqrt{x+1}+3=0$

Solution:

$\begin{aligned} & x-3 \sqrt{x+1}+3=0 \\ & \text { Let } \sqrt{x+1}=t \\ & \Rightarrow x+1=t^2 \\ & \Rightarrow x=t^2-1\end{aligned}$

So, the equation becomes

$\begin{aligned} & \left(t^2-1\right)-3 t+3=0 \\ \Rightarrow & t^2-3 t+2=0 \\ \Rightarrow & (t-1)(t-2)=0 \\ \Rightarrow & t=1, t=2 \\ \Rightarrow & \sqrt{x+1}=1, \quad \sqrt{x+1}=2 \\ \Rightarrow & x+1=1, \quad x+1=4 \\ \Rightarrow & x=0, \quad x=3\end{aligned}$