2.5 Equations and inequalities

Detailed Theory: Equations and Inequalities

1. Basic Concepts of Equations

1.1 What is an Equation?

An equation is a mathematical statement that asserts the equality of two expressions.

General Form: $LHS = RHS$

where LHS = Left Hand Side, RHS = Right Hand Side

Example: $2x + 3 = 7$

Here, LHS = $2x + 3$, RHS = $7$

1.2 Solution/Root of an Equation

A value of the variable that makes the equation true is called a solution or root.

Example: For $2x + 3 = 7$, substituting $x = 2$:

LHS = $2(2) + 3 = 4 + 3 = 7$ = RHS

So $x = 2$ is a solution.

1.3 Types of Equations

a) Identity

An equation that is true for all values of the variable.

Example: $(x+1)^2 = x^2 + 2x + 1$

This is true for all real $x$.

b) Conditional Equation

An equation that is true only for specific values of the variable.

Example: $2x + 3 = 7$ is true only when $x = 2$

c) Inconsistent Equation

An equation that has no solution.

Example: $x + 2 = x + 5$

Subtracting $x$ from both sides: $2 = 5$ (false)

So no solution exists.

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2. Linear Equations

2.1 Standard Form

A linear equation in one variable has the form:

$ax + b = 0$ where $a \neq 0$

Solution: $x = -\frac{b}{a}$

2.2 Solving Linear Equations

Basic Principle: Perform the same operation on both sides to isolate the variable.

Steps:

1. Simplify both sides (remove parentheses, combine like terms)

2. Collect variable terms on one side, constants on the other

3. Solve for the variable

Example: Solve $3(x-2) + 5 = 2x + 1$

Step 1: Simplify: $3x - 6 + 5 = 2x + 1$ $3x - 1 = 2x + 1$

Step 2: Collect terms: $3x - 2x = 1 + 1$ $x = 2$

Step 3: Check: LHS = $3(2-2) + 5 = 3(0) + 5 = 5$ RHS = $2(2) + 1 = 4 + 1 = 5$ Equal, so $x=2$ is correct.

2.3 Linear Equations in Two Variables

Standard form: $ax + by + c = 0$ where $a$ and $b$ are not both zero.

Solution: Infinitely many ordered pairs $(x, y)$ satisfy the equation.

Graph: Always a straight line.

2.4 System of Linear Equations

Two or more linear equations considered together.

a) System of Two Equations

$a_1x + b_1y = c_1$ $a_2x + b_2y = c_2$

Possible Solutions:

1. Unique solution: Lines intersect at one point

2. No solution: Lines are parallel

3. Infinitely many solutions: Lines coincide

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3. Quadratic Equations

3.1 Standard Form

A quadratic equation in one variable has the form:

$ax^2 + bx + c = 0$ where $a \neq 0$

3.2 Solving Quadratic Equations

a) Factorization Method

Write $ax^2 + bx + c$ as product of two linear factors.

Example: Solve $x^2 - 5x + 6 = 0$

Factor: $(x-2)(x-3) = 0$

So $x-2=0$ or $x-3=0$

Solutions: $x=2$, $x=3$

b) Completing the Square

Convert to form $(x-p)^2 = q$

Example: Solve $x^2 - 6x + 5 = 0$

Step 1: Move constant: $x^2 - 6x = -5$

Step 2: Add square of half coefficient of $x$: $(-6/2)^2 = 9$

$x^2 - 6x + 9 = -5 + 9$

$(x-3)^2 = 4$

Step 3: Take square root: $x-3 = \pm 2$

So $x = 3 \pm 2$

Solutions: $x=5$, $x=1$

c) Quadratic Formula

The solutions of $ax^2 + bx + c = 0$ are:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

3.3 Discriminant

For $ax^2 + bx + c = 0$, discriminant $D = b^2 - 4ac$

Nature of Roots:

1. $D > 0$: Two distinct real roots

2. $D = 0$: One real root (repeated/double root)

3. $D < 0$: Two complex conjugate roots

3.4 Sum and Product of Roots

For $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$:

Sum of roots: $\alpha + \beta = -\frac{b}{a}$

Product of roots: $\alpha\beta = \frac{c}{a}$

3.5 Formation of Quadratic Equation

If roots are $\alpha$ and $\beta$, the quadratic equation is:

$x^2 - (\alpha+\beta)x + \alpha\beta = 0$

Example: Form quadratic with roots $2$ and $-3$

Sum = $2 + (-3) = -1$

Product = $2 \times (-3) = -6$

Equation: $x^2 - (-1)x + (-6) = 0$

$x^2 + x - 6 = 0$

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4. Cubic Equations

4.1 Standard Form

$ax^3 + bx^2 + cx + d = 0$ where $a \neq 0$

4.2 Solving Cubic Equations

a) Factorization Method

Look for rational roots using Rational Root Theorem.

Example: Solve $x^3 - 6x^2 + 11x - 6 = 0$

Try factors of $-6$: $\pm1, \pm2, \pm3, \pm6$

Check: $P(1) = 1 - 6 + 11 - 6 = 0 \quad \checkmark$

So $(x-1)$ is a factor.

Divide: $(x^3 - 6x^2 + 11x - 6) \div (x-1) = x^2 - 5x + 6$

So $(x-1)(x^2 - 5x + 6) = 0$

$(x-1)(x-2)(x-3) = 0$

Roots: $x=1, 2, 3$

b) Sum and Product of Roots

For $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha, \beta, \gamma$:

1. $\alpha + \beta + \gamma = -\frac{b}{a}$

2. $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$

3. $\alpha\beta\gamma = -\frac{d}{a}$

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5. Rational Equations

5.1 Definition

An equation containing rational expressions (fractions with variables in denominator).

Example: $\frac{x}{x-2} + \frac{3}{x+1} = 5$

5.2 Solving Rational Equations

Steps:

1. Find LCD (Least Common Denominator)

2. Multiply both sides by LCD to eliminate fractions

3. Solve resulting equation

4. Check solutions don't make any denominator zero

Example: Solve $\frac{1}{x} + \frac{1}{x+2} = \frac{5}{6}$

Step 1: LCD = $x(x+2) \times 6 = 6x(x+2)$

Step 2: Multiply by LCD:

$6(x+2) + 6x = 5x(x+2)$

$6x + 12 + 6x = 5x^2 + 10x$

Step 3: Simplify: $12x + 12 = 5x^2 + 10x$

$0 = 5x^2 - 2x - 12$

$5x^2 - 2x - 12 = 0$

Use quadratic formula:

$x = \frac{2 \pm \sqrt{4 + 240}}{10} = \frac{2 \pm \sqrt{244}}{10} = \frac{2 \pm 2\sqrt{61}}{10} = \frac{1 \pm \sqrt{61}}{5}$

Step 4: Check denominators: Neither solution makes $x=0$ or $x=-2$

So solutions: $x = \frac{1 + \sqrt{61}}{5}$ and $x = \frac{1 - \sqrt{61}}{5}$

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6. Radical Equations

6.1 Definition

Equations containing variables under radical signs (square roots, cube roots, etc.).

Example: $\sqrt{x+3} = x - 3$

6.2 Solving Radical Equations

Steps:

1. Isolate one radical

2. Raise both sides to appropriate power to eliminate radical

3. Solve resulting equation

4. CRUCIAL: Check all solutions in original equation (extraneous solutions may appear)

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

Step 1: Radical already isolated

Step 2: Square both sides:

$(\sqrt{x+3})^2 = (x-3)^2$

$x + 3 = x^2 - 6x + 9$

Step 3: Rearrange: $0 = x^2 - 7x + 6$

Factor: $(x-1)(x-6) = 0$

Potential solutions: $x=1$, $x=6$

Step 4: Check in original equation:

For $x=1$: LHS = $\sqrt{1+3} = \sqrt{4} = 2$ RHS = $1-3 = -2$ ✗ (not equal)

For $x=6$: LHS = $\sqrt{6+3} = \sqrt{9} = 3$ RHS = $6-3 = 3$ ✓

So only solution is $x=6$

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7. Exponential Equations

7.1 Definition

Equations where the variable appears in exponents.

Example: $2^x = 8$

7.2 Solving Exponential Equations

a) Same Base Method

If $a^x = a^y$, then $x = y$ (for $a > 0$, $a \neq 1$)

Example: Solve $3^{2x-1} = 27$

Write $27 = 3^3$:

$3^{2x-1} = 3^3$

So $2x-1 = 3$

$2x = 4$

$x = 2$

b) Using Logarithms

If $a^x = b$, then $x = \frac{\log b}{\log a}$

Example: Solve $2^x = 5$

Take log: $x\log 2 = \log 5$

$x = \frac{\log 5}{\log 2} \approx \frac{0.6990}{0.3010} \approx 2.322$

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8. Logarithmic Equations

8.1 Definition

Equations containing logarithms of expressions involving the variable.

Example: $\log(x+2) + \log(x-1) = 1$

8.2 Solving Logarithmic Equations

Steps:

1. Use logarithm properties to combine/simplify

2. Convert to exponential form

3. Solve resulting equation

4. Check solutions satisfy domain restrictions (arguments must be positive)

Example: Solve $\log(x+2) + \log(x-1) = 1$

Step 1: Combine using product rule: $\log[(x+2)(x-1)] = 1$

Step 2: Convert to exponential: $(x+2)(x-1) = 10^1 = 10$

Step 3: Expand: $x^2 + x - 2 = 10$

$x^2 + x - 12 = 0$

Factor: $(x+4)(x-3) = 0$

Potential solutions: $x=-4$, $x=3$

Step 4: Check domain:

For $x=-4$: $x+2=-2<0$ ✗ (invalid)

For $x=3$: $x+2=5>0$, $x-1=2>0$ ✓

So solution is $x=3$

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9. Absolute Value Equations

9.1 Definition

Equations containing absolute value expressions.

Example: $|2x-3| = 7$

9.2 Solving Absolute Value Equations

For $|f(x)| = a$ where $a \geq 0$:

Case 1: $f(x) = a$ Case 2: $f(x) = -a$

Example: Solve $|2x-3| = 7$

Case 1: $2x-3 = 7$ $2x = 10$ $x = 5$

Case 2: $2x-3 = -7$ $2x = -4$ $x = -2$

Solutions: $x=5$, $x=-2$

Check: $|2(5)-3| = |10-3| = |7| = 7$ ✓ $|2(-2)-3| = |-4-3| = |-7| = 7$ ✓

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10. Inequalities - Basic Concepts

10.1 What is an Inequality?

A mathematical statement comparing two expressions using inequality symbols:

• $<$: less than

• $\leq$: less than or equal to

• $>$: greater than

• $\geq$: greater than or equal to

Example: $2x + 3 < 7$

10.2 Properties of Inequalities

a) Addition/Subtraction

If $a < b$, then $a + c < b + c$ for any real $c$

If $a < b$, then $a - c < b - c$ for any real $c$

b) Multiplication/Division by Positive Number

If $a < b$ and $c > 0$, then $ac < bc$ and $\frac{a}{c} < \frac{b}{c}$

c) Multiplication/Division by Negative Number

If $a < b$ and $c < 0$, then $ac > bc$ and $\frac{a}{c} > \frac{b}{c}$

IMPORTANT: Inequality sign reverses when multiplying/dividing by negative number!

10.3 Solution of Inequalities

The set of all values that satisfy the inequality.

Example: Solve $2x + 3 < 7$

Subtract 3: $2x < 4$

Divide by 2: $x < 2$

Solution: All real numbers less than 2.

In interval notation: $(-\infty, 2)$

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11. Linear Inequalities

11.1 Solving Linear Inequalities

Similar to solving linear equations, but remember to reverse inequality when multiplying/dividing by negative.

Example 1: Solve $3x - 2 \leq 7$

Add 2: $3x \leq 9$

Divide by 3: $x \leq 3$

Solution: $x \in (-\infty, 3]$

Example 2: Solve $-2x + 5 > 1$

Subtract 5: $-2x > -4$

Divide by -2 (reverse inequality): $x < 2$

Solution: $x \in (-\infty, 2)$

11.2 Compound Inequalities

Two inequalities connected by "and" or "or".

a) "And" Inequalities (Conjunction)

Both inequalities must be true.

Example: Solve $-3 \leq 2x + 1 < 5$

This means: $-3 \leq 2x + 1$ AND $2x + 1 < 5$

Solve simultaneously:

Subtract 1 from all parts: $-4 \leq 2x < 4$

Divide by 2: $-2 \leq x < 2$

Solution: $x \in [-2, 2)$

b) "Or" Inequalities (Disjunction)

At least one inequality must be true.

Example: Solve $2x - 3 < 1$ OR $x + 4 > 7$

Solve separately:

First: $2x - 3 < 1 \Rightarrow 2x < 4 \Rightarrow x < 2$

Second: $x + 4 > 7 \Rightarrow x > 3$

Since it's "OR", solution is: $x < 2$ OR $x > 3$

In interval notation: $(-\infty, 2) \cup (3, \infty)$

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12. Quadratic Inequalities

12.1 Solving Quadratic Inequalities

Method:

1. Solve corresponding quadratic equation $ax^2 + bx + c = 0$

2. Use sign chart/test intervals

3. Determine solution based on inequality

Example 1: Solve $x^2 - 3x - 4 > 0$

Step 1: Solve $x^2 - 3x - 4 = 0$

Factor: $(x-4)(x+1) = 0$

Roots: $x = -1$, $x = 4$

Step 2: These divide number line into 3 intervals:

1. $x < -1$

2. $-1 < x < 4$

3. $x > 4$

Test sign of $(x-4)(x+1)$ in each interval:

Interval 1 (test $x=-2$): $(-)(-) = +$ (positive)

Interval 2 (test $x=0$): $(-)(+) = -$ (negative)

Interval 3 (test $x=5$): $(+)(+) = +$ (positive)

Step 3: We want $> 0$ (positive), so solution is:

$x < -1$ OR $x > 4$

In interval notation: $(-\infty, -1) \cup (4, \infty)$

Example 2: Solve $x^2 - 4x + 4 \leq 0$

Step 1: Solve $x^2 - 4x + 4 = 0$

$(x-2)^2 = 0$

Root: $x = 2$ (double root)

Step 2: Test intervals:

Since $(x-2)^2 \geq 0$ always, and equals 0 only at $x=2$

Step 3: We want $\leq 0$, so solution is only $x=2$

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13. Rational Inequalities

13.1 Solving Rational Inequalities

Method:

1. Bring all terms to one side (get 0 on other side)

2. Combine into single fraction

3. Find critical points (zeros of numerator and denominator)

4. Use sign chart/test intervals

5. Check endpoints based on inequality ($\leq$ or $<$)

Example: Solve $\frac{x-1}{x+2} \geq 0$

Step 1: Already in correct form

Step 2: Critical points:

Numerator zero: $x-1=0 \Rightarrow x=1$

Denominator zero: $x+2=0 \Rightarrow x=-2$

Step 3: These divide number line into 3 intervals:

1. $x < -2$

2. $-2 < x < 1$

3. $x > 1$

Step 4: Test sign of $\frac{x-1}{x+2}$:

Interval 1 (test $x=-3$): $\frac{-}{-} = +$ (positive)

Interval 2 (test $x=0$): $\frac{-}{+} = -$ (negative)

Interval 3 (test $x=2$): $\frac{+}{+} = +$ (positive)

Step 5: We want $\geq 0$ (positive or zero):

At $x=1$: fraction = 0 ✓ (include)

At $x=-2$: undefined ✗ (exclude)

Solution: $x < -2$ OR $x \geq 1$

In interval notation: $(-\infty, -2) \cup [1, \infty)$

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14. Absolute Value Inequalities

14.1 Solving Absolute Value Inequalities

Rule 1: $|f(x)| < a$ where $a > 0$ is equivalent to: $-a < f(x) < a$

Rule 2: $|f(x)| > a$ where $a > 0$ is equivalent to: $f(x) < -a$ OR $f(x) > a$

Example 1: Solve $|2x-3| < 7$

By Rule 1: $-7 < 2x-3 < 7$

Add 3: $-4 < 2x < 10$

Divide by 2: $-2 < x < 5$

Solution: $x \in (-2, 5)$

Example 2: Solve $|3x+2| \geq 5$

By Rule 2: $3x+2 \leq -5$ OR $3x+2 \geq 5$

Solve first: $3x \leq -7 \Rightarrow x \leq -\frac{7}{3}$

Solve second: $3x \geq 3 \Rightarrow x \geq 1$

Solution: $x \leq -\frac{7}{3}$ OR $x \geq 1$

In interval notation: $(-\infty, -\frac{7}{3}] \cup [1, \infty)$

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15. Systems of Inequalities

15.1 Solving Systems Graphically

The solution is the intersection of all individual solution sets.

Example: Solve system: $y > 2x - 1$ $y \leq -x + 3$

Graph each inequality:

1. $y > 2x - 1$: Dotted line $y = 2x - 1$, shade above

2. $y \leq -x + 3$: Solid line $y = -x + 3$, shade below

Solution is the overlapping shaded region.

15.2 Solving Algebraically

Find intersection of solution sets.

Example: Solve system: $x + 2y \leq 6$ $3x - y \geq 2$

Solve each for $y$:

First: $2y \leq 6 - x \Rightarrow y \leq 3 - \frac{x}{2}$

Second: $-y \geq 2 - 3x \Rightarrow y \leq 3x - 2$ (reverse inequality!)

So we need: $y \leq 3 - \frac{x}{2}$ AND $y \leq 3x - 2$

The solution is all $(x,y)$ satisfying both conditions.

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16. Special Equations and Inequalities

16.1 Equations Involving Modulus and Squares

Important Identity: $|x|^2 = x^2$

Example: Solve $|x-2| = |2x+1|$

Square both sides: $(x-2)^2 = (2x+1)^2$

Expand: $x^2 - 4x + 4 = 4x^2 + 4x + 1$

Rearrange: $0 = 3x^2 + 8x - 3$

Use quadratic formula: $x = \frac{-8 \pm \sqrt{64 + 36}}{6} = \frac{-8 \pm \sqrt{100}}{6} = \frac{-8 \pm 10}{6}$

So $x = \frac{2}{6} = \frac{1}{3}$ or $x = \frac{-18}{6} = -3$

Check both satisfy original equation.

16.2 Reciprocal Equations

Equations of form: $x + \frac{1}{x} = a$

Solution: Multiply by $x$: $x^2 + 1 = ax$

$x^2 - ax + 1 = 0$

Use quadratic formula: $x = \frac{a \pm \sqrt{a^2 - 4}}{2}$

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17. Word Problems with Equations

17.1 Age Problems

Example: John is 5 years older than Mary. In 3 years, John will be twice as old as Mary. Find their current ages.

Solution: Let Mary's age = $x$ Then John's age = $x+5$

In 3 years: Mary's age = $x+3$ John's age = $(x+5)+3 = x+8$

Given: $x+8 = 2(x+3)$

Solve: $x+8 = 2x+6$

$8-6 = 2x-x$

$2 = x$

So Mary is 2, John is 7.

Check: In 3 years, Mary=5, John=10 (John is twice as old) ✓

17.2 Mixture Problems

Example: How many liters of 20% salt solution must be added to 10 liters of 50% salt solution to get 30% salt solution?

Solution: Let $x$ = liters of 20% solution

Salt in 20% solution = $0.20x$ Salt in 50% solution = $0.50 \times 10 = 5$ Total salt = $0.20x + 5$

Total solution = $x + 10$

Concentration = $\frac{0.20x + 5}{x + 10} = 0.30$

Solve: $0.20x + 5 = 0.30(x + 10)$

$0.20x + 5 = 0.30x + 3$

$5-3 = 0.30x - 0.20x$

$2 = 0.10x$

$x = 20$ liters

17.3 Work/Rate Problems

Example: Pipe A fills a tank in 4 hours, Pipe B fills it in 6 hours. How long to fill using both pipes together?

Solution: Rate of A = $\frac{1}{4}$ tank/hour Rate of B = $\frac{1}{6}$ tank/hour

Combined rate = $\frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}$ tank/hour

Time = $\frac{1}{\text{rate}} = \frac{1}{5/12} = \frac{12}{5} = 2.4$ hours = 2 hours 24 minutes

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18. Important Formulas and Summary

18.1 Quadratic Formula

For $ax^2 + bx + c = 0$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

18.2 Discriminant

$D = b^2 - 4ac$

18.3 Sum and Product of Roots

For $ax^2 + bx + c = 0$ with roots $\alpha, \beta$:

$\alpha + \beta = -\frac{b}{a}$

$\alpha\beta = \frac{c}{a}$

18.4 Absolute Value Rules

1. $|x| = a \Rightarrow x = a$ or $x = -a$ (for $a \geq 0$)

2. $|x| < a \Rightarrow -a < x < a$ (for $a > 0$)

3. $|x| > a \Rightarrow x < -a$ or $x > a$ (for $a > 0$)

18.5 Inequality Reversal

If $a < b$ and $c < 0$, then: $ac > bc$ and $\frac{a}{c} > \frac{b}{c}$

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19. Exam Tips and Common Mistakes

19.1 Common Mistakes

1. Not checking solutions: Always check in original equation, especially for rational and radical equations

2. Forgetting to reverse inequality sign when multiplying/dividing by negative

3. Extraneous solutions: Solutions that don't satisfy original equation (common in radical and rational equations)

4. Domain restrictions: For logarithms (argument > 0), fractions (denominator ≠ 0)

5. Absolute value: Remember both positive and negative cases

19.2 Problem-Solving Strategy

1. Identify the type: Linear, quadratic, rational, radical, etc.

2. Choose appropriate method: Factorization, quadratic formula, substitution, etc.

3. Solve systematically: Show all steps

4. Check solutions: Verify in original equation

5. Write final answer clearly: In simplest form, with appropriate notation

19.3 Quick Checks

1. Linear equations: Degree 1, graph is straight line

2. Quadratic equations: Degree 2, can have 0, 1, or 2 real solutions

3. Rational equations: Check denominator ≠ 0

4. Radical equations: Check for extraneous solutions

5. Inequalities: Reverse sign when multiplying/dividing by negative

This comprehensive theory covers all aspects of equations and inequalities with detailed explanations and examples, providing complete preparation for the entrance examination.