Irrational equations and Inequalities: Problems with Solutions

What are Irrational Equations?

Understand the basics of irrational equations, involving roots or surds in variables. This section covers key forms, solution steps, and conditions for solving irrational inequalities and equations effectively.

Definition of Irrational Equations

An irrational equation is an algebraic equation that contains one or more irrational expressions, usually involving square roots, cube roots, or higher-order roots of algebraic expressions. These equations often require careful manipulation to isolate and eliminate the irrational part.

For example, an equation like
$\sqrt{x + 2} = x - 1$ is an irrational equation because the variable $x$ is under a square root.

Common Forms and Examples

Some common forms of irrational equations include:

• $\sqrt{f(x)} = g(x)$

• $\sqrt{f(x)} + \sqrt{g(x)} = h(x)$

• $\frac{1}{\sqrt{x}} = x + 1$

Examples:

1. $\sqrt{2x + 3} = x - 1$

2. $\sqrt{x + 4} + \sqrt{3x - 1} = 5$

3. $\frac{1}{\sqrt{x - 2}} = x$

These forms frequently form the base for irrational inequality questions as well, when an inequality sign is involved.

Conditions for Existence of Solutions

Before solving irrational equations, it is important to identify the domain of validity. Since square roots (and other even-order roots) are defined only for non-negative values, we must ensure:

• The expression inside a square root must be non-negative, i.e.,
  If $\sqrt{f(x)}$ appears, then solve $f(x) \geq 0$

• If the irrational term appears in a denominator, then the expression inside the root must be strictly positive, i.e.,
  If $\frac{1}{\sqrt{f(x)}}$ appears, then solve $f(x) > 0$

These domain restrictions play a crucial role while solving irrational inequalities and must not be ignored.

Steps to Solve Irrational Equations

Here’s a general step-by-step strategy to solve an irrational equation:

1. Isolate the Irrational Expression:
   Try to bring the root term (like $\sqrt{f(x)}$) to one side of the equation.

2. Square Both Sides:
   Remove the square root by squaring both sides. For example,
   $\sqrt{x + 2} = x - 1 \Rightarrow x + 2 = (x - 1)^2$

3. Simplify and Solve the Resulting Polynomial Equation:
   After removing the root, simplify the expression and solve for $x$.

4. Check for Extraneous Solutions:
   Squaring may introduce invalid solutions. Substitute your answers back into the original equation to verify their validity.

What are Irrational Inequalities?

Understand the basics of irrational equations, involving roots or surds in variables. This section covers key forms, solution steps, and conditions for solving irrational inequalities and equations effectively.

Definition of Irrational Inequalities

An irrational inequality is an inequality that includes at least one irrational expression, usually involving roots like square roots or cube roots of algebraic functions. These types of inequalities require careful handling due to the presence of non-linear terms.

For example:

• $\sqrt{x - 1} > 3$

• $\frac{1}{\sqrt{x + 2}} < 4$

• $\sqrt{2x + 3} \leq x - 1$

Such inequalities are frequently tested in irrational inequalities Class 11 and also form an integral part of the irrational inequalities JEE syllabus.

Forms of Inequalities:

Common forms of irrational inequalities include:

1. $\sqrt{f(x)} > g(x)$

2. $\sqrt{f(x)} < g(x)$

3. $\frac{1}{\sqrt{f(x)}} \geq g(x)$

4. $\sqrt{f(x)} + \sqrt{g(x)} < h(x)$

Each type requires a specific approach that involves squaring, domain analysis, and sign analysis.

Inequalities are the relationship between two expressions that are not equal to one another. Symbols denoting the inequalities are $<,\ >,\ \leq,\ \geq$ and $\neq$.

• $x<4$, "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 that of equality, but instead of the 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: $|x-2| \leq 5$

Key Concepts to Consider

Before solving irrational inequalities, it’s important to remember the following:

1. Domain Restrictions:
   If the inequality involves $\sqrt{f(x)}$, then the domain is given by $f(x) \geq 0$.

2. Squaring Caution:
   Squaring both sides of an inequality can change the nature of the inequality if signs are not properly accounted for. It is valid only when both sides are non-negative.

3. Critical Points and Intervals:
   After simplification, always break the solution into intervals using critical points. Test each interval in the original inequality.

4. Reject Extraneous Solutions:
   As with irrational equations, squaring may introduce false roots. Always verify with the original inequality.

Example Problems

Solve: $\sqrt{x + 1} \leq x$

Step 1: Domain

• $x + 1 \geq 0 \Rightarrow x \geq -1$

Step 2: Square both sides

• $\sqrt{x + 1} \leq x \Rightarrow x + 1 \leq x^2$

Step 3: Rearrange

• $x^2 - x - 1 \geq 0$

Step 4: Solve the quadratic inequality

• The roots of $x^2 - x - 1 = 0$ are $x = \frac{1 \pm \sqrt{5}}{2}$

Step 5: Find intervals satisfying the inequality

• Use sign analysis to find that $x \leq \frac{1 - \sqrt{5}}{2}$ or $x \geq \frac{1 + \sqrt{5}}{2}$
  But from domain, $x \geq -1$

Final Answer:
$x \in \left[\frac{1 + \sqrt{5}}{2}, \infty\right)$

Graphical Interpretation

For e.g. a $>3$ gives us a range of solutions, means $(3, \infty)$
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 inequalities are inequalities with irrational equations. For example, $\sqrt{x^2+7} = 4$

If $n$ is odd

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 the 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 not be less than RHS for all such values
c. If $g(x) \geq 0$, then solve $f(x)<(g(x))^n$

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

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 the RHS cannot be greater than the 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 inequality

$6-2 x<0$

$\Rightarrow x>2$

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 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 sides

$
\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 sides

$
\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:

Roots of the 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

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

$ -2t+6+t^2-6t+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 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

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

List of Topics Related to Irrational Equations and Inequalities

Explore the most relevant topics connected to irrational equations and various classes of inequalities. Each link opens a Careers360 topic page with explanations, formulas, and solved examples.

NCERT Resources

Explore curated NCERT study materials for Class 11 Chapter 5 "Complex Numbers and Quadratic Equations" including detailed notes, solutions, and exemplar problems. These resources are essential for mastering the fundamentals and aligning with the CBSE curriculum.

Practice Questions Based on Irrational Equations and Inequalities

Sharpen your skills with irrational inequalities questions and irrational equation problems through concept-based MCQs. These practice sets are designed to strengthen your understanding of domain analysis, squaring techniques, and solution verification.

Irrational Equations and Inequalities - Practice Question MCQ

Below are the next topics, which you can practice to gain more understanding: