Algebraic Equations: Definition, Types, Formulas & Solved Examples

What Is an Algebraic Equation?

An algebraic equation is a statement that two algebraic expressions are equal. That's it. Strip away the jargon and you get: something on the left side of an "=" sign has the same value as something on the right side.

$$P = Q$$

where $P$ and $Q$ are expressions built from variables, constants, and arithmetic operations.

Take $2x + 3 = 11$. The expression $2x + 3$ equals $11$. Your job is to find the value of $x$ that keeps both sides balanced.

Here's the part most textbooks skip: an equation is fundamentally a question. When you write $2x + 3 = 11$, you're really asking, "What number, when doubled and increased by 3, gives 11?" Thinking of equations as questions — not as things to mechanically "solve" — changes how students engage with algebra entirely.

Algebraic Equation vs Algebraic Expression - The Confusion That Costs Marks

About 4 out of 10 students we work with at Bhanzu mix up expressions and equations in their first algebra session. The difference is simple but critical:

An expression is like an incomplete sentence. An equation is the full statement. You can simplify $3x + 7$, but you can't "solve" it — there's nothing to solve for until you set it equal to something.

Anatomy of an Algebraic Equation

Let's break down $5x^2 - 3x + 7 = 0$ piece by piece:

• Variables: $x$ — the unknowns you're solving for.

• Coefficients: $5$ and $-3$ — the numbers multiplied by the variables.

• Constant: $7$ — the standalone number with no variable attached.

• Degree: $2$ — the highest exponent on any variable. This tells you the type of equation and how many solutions to expect.

• Operator: The $=$ sign. Everything before it is the Left-Hand Side (LHS), everything after is the Right-Hand Side (RHS).

Quick mental check: If someone writes $5x + 9 > 12$, is that an equation? No. The ">" makes it an inequality. Equations exclusively use the $=$ sign.

Types of Algebraic Equations

The type of an algebraic equation depends on its degree — the highest power of the variable. The degree tells you two things: what the equation is called, and the maximum number of solutions it can have.

1. Linear Algebraic Equations (Degree 1)

A linear equation has the variable raised to the power of 1. No squares, no cubes — just the variable sitting there plainly.

General form:

$$ax + b = 0 \quad (a \neq 0)$$

Examples:

• $3x + 5 = 20$

• $7y - 14 = 0$

• $2x + 3y = 12$ (linear equation in two variables)

Linear equations are the first real "algebra" students encounter, typically in Class 7 (NCERT Chapter 4) or Grade 7 (CCSS 7.EE.1). And they're where the most fundamental solving habit — doing the same thing to both sides — either gets built or doesn't.

What makes linear equations "linear"? If you plot every solution of a linear equation on a graph, you get a straight line. That's literally where the name comes from. The equation $y = 2x + 1$ draws a perfect line through the coordinate plane — no curves, no bends.

One variable vs two variables:

2. Quadratic Algebraic Equations (Degree 2)

The variable gets squared. That single change — going from $x$ to $x^2$ — transforms everything about how the equation behaves.

General form:

$$ax^2 + bx + c = 0 \quad (a \neq 0)$$

Why $a \neq 0$? Because if $a = 0$, the $x^2$ term vanishes and you're back to a linear equation. The $a$ is what makes it quadratic.

Examples:

• $x^2 - 5x + 6 = 0$

• $2x^2 + 3x - 2 = 0$

• $x^2 = 49$ (yes, this counts — rewrite it as $x^2 - 49 = 0$)

Quadratic equations can have zero, one, or two real solutions. The discriminant $\Delta = b^2 - 4ac$ tells you which scenario you're in before you even solve:

The mistake students make most often: Writing $x^2 = 49$ and concluding $x = 7$. They forget the negative root. The correct answer is $x = 7$ or $x = -7$. In our experience, roughly half of students lose marks on this in their first quadratic test — not because they don't know the concept, but because they stop thinking too early.

3. Cubic Algebraic Equations (Degree 3)

General form:

$$ax^3 + bx^2 + cx + d = 0 \quad (a \neq 0)$$

Cubic equations can have up to three real roots. Unlike quadratics, a cubic always has at least one real root — the graph of a cubic function must cross the x-axis at least once.

Examples:

• $x^3 - 6x^2 + 11x - 6 = 0$

• $2x^3 + 3x^2 - 1 = 0$

Students typically encounter cubics in Class 11 (NCERT) or in competition math. The factoring approach is usually the first method taught: check whether simple values like $x = 1, -1, 2, -2$ satisfy the equation (using the Factor Theorem), then reduce the cubic to a quadratic.

4. Polynomial Algebraic Equations (Higher Degree)

Any algebraic equation of degree 4 or higher falls into this category.

$$a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0$$

Quartic example (degree 4): $x^4 - 5x^2 + 4 = 0$

This particular quartic is solvable by substitution — let $u = x^2$, and it becomes $u^2 - 5u + 4 = 0$, a quadratic. Not all higher-degree equations are this friendly, though. In fact, there's a proven mathematical result (the Abel–Ruffini theorem) showing that general equations of degree 5 or higher have no formula using basic operations. That's not a limitation of our knowledge — it's a hard mathematical fact.

5. Rational Algebraic Equations

These have variables in the denominator.

$$\frac{P(x)}{Q(x)} = 0$$

Example:

$$\frac{x^2 - 4}{x - 1} = 0$$

The key rule: the denominator $Q(x)$ can never equal zero. So here, $x \neq 1$. Solve the numerator $x^2 - 4 = 0$ to get $x = 2$ or $x = -2$, then verify neither makes the denominator zero. Both are valid here.

The trap with rational equations: Students solve the numerator, get valid-looking answers, and forget to check the denominator. On an exam, this is a silent mark-killer.

Algebraic Equations Formulas

These are the identities and formulas you'll use repeatedly when working with algebraic equations. Memorising them saves time; understanding them prevents errors.

Essential Algebraic Identities

$$ (a + b)^2 = a^2 + 2ab + b^2 $$

$$ (a - b)^2 = a^2 - 2ab + b^2 $$

$$ a^2 - b^2 = (a + b)(a - b) $$

$$ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 $$

$$ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $$

$$ a^3 + b^3 = (a + b)(a^2 - ab + b^2) $$

$$ a^3 - b^3 = (a - b)(a^2 + ab + b^2) $$

$$ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca $$

The Quadratic Formula

For any quadratic equation $ax^2 + bx + c = 0$:

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

This formula works every single time — even when factoring doesn't. It's the universal fallback for degree-2 equations.

Where the $\pm$ does the heavy lifting: The $\pm$ symbol means you calculate twice — once with $+$, once with $-$. That's why quadratics can have two roots.

Discriminant

$$ \Delta = b^2 - 4ac $$

Before plugging into the full quadratic formula, compute the discriminant. It tells you whether to expect two roots, one root, or no real roots (see the table in the quadratic section above).

Sum and Product of Roots (Vieta's Formulas for Quadratics)

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

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

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

These are incredibly useful for checking your answers quickly. Solved a quadratic and got roots $2$ and $3$? Sum should be $5$, product should be $6$. If the equation is $x^2 - 5x + 6 = 0$, that checks out: $-(-5)/1 = 5$ and $6/1 = 6$.

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

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

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

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

How to Solve Algebraic Equations - Step by Step

Different equation types need different methods. But one golden rule applies to every single one: whatever you do to one side, you must do to the other.

Solving Linear Equations

Method: Isolate the variable using inverse operations.

Example 1: Solve $3x + 7 = 22$

$$3x + 7 = 22$$

Subtract 7 from both sides:

$$3x = 22 - 7$$

$$3x = 15$$

Divide both sides by 3:

$$x = \frac{15}{3} = 5$$

Verification: $3(5) + 7 = 15 + 7 = 22$ ✓

Example 2: Solve $\frac{2x - 5}{3} = 7$

Multiply both sides by 3:

$$2x - 5 = 21$$

Add 5 to both sides:

$$2x = 26$$

$$x = 13$$

Verification: $\frac{2(13) - 5}{3} = \frac{26 - 5}{3} = \frac{21}{3} = 7$ ✓

Example 3 (Variables on both sides): Solve $5x - 9 = 2x + 12$

Move all $x$ terms to the left:

$$5x - 2x = 12 + 9$$

$$3x = 21$$

$$x = 7$$

Common student error here: When moving $-9$ from LHS to RHS, students write $12 - 9$ instead of $12 + 9$. The sign flips when a term crosses the equals sign. This single mistake accounts for more lost marks in Class 7–8 algebra than any other.

Solving Quadratic Equations

Three methods. Use the one that fits the equation best.

Method 1: Factoring

Works when the quadratic factors neatly into two binomials.

Example 4: Solve $x^2 - 7x + 12 = 0$

Find two numbers that multiply to $12$ and add to $-7$. Those numbers are $-3$ and $-4$.

$$(x - 3)(x - 4) = 0$$

By the zero-product property (if $A \times B = 0$, then $A = 0$ or $B = 0$):

$$x - 3 = 0 \quad \Rightarrow \quad x = 3$$

$$x - 4 = 0 \quad \Rightarrow \quad x = 4$$

Verification:

• $x = 3$: $(3)^2 - 7(3) + 12 = 9 - 21 + 12 = 0$ ✓

• $x = 4$: $(4)^2 - 7(4) + 12 = 16 - 28 + 12 = 0$ ✓

Method 2: Quadratic Formula

Works for every quadratic equation — use this when factoring isn't obvious.

Example 5: Solve $2x^2 + 5x - 3 = 0$

Here $a = 2$, $b = 5$, $c = -3$.

$$\Delta = b^2 - 4ac = 25 - 4(2)(-3) = 25 + 24 = 49$$

$$x = \frac{-5 \pm \sqrt{49}}{2 \times 2} = \frac{-5 \pm 7}{4}$$

$$x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}$$

$$x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3$$

Solutions: $x = \frac{1}{2}$ or $x = -3$

Quick check with Vieta's formulas:

• Sum of roots: $\frac{1}{2} + (-3) = -\frac{5}{2} = -\frac{b}{a} = -\frac{5}{2}$ ✓

• Product of roots: $\frac{1}{2} \times (-3) = -\frac{3}{2} = \frac{c}{a} = \frac{-3}{2}$ ✓

Method 3: Completing the Square

Useful when you need the vertex form or when dealing with equations that don't factor cleanly.

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

Move the constant:

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

Take half the coefficient of $x$ (which is $6 \div 2 = 3$), square it ($3^2 = 9$), and add to both sides:

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

$$(x + 3)^2 = 7$$

$$x + 3 = \pm\sqrt{7}$$

$$x = -3 + \sqrt{7} \quad \text{or} \quad x = -3 - \sqrt{7}$$

$$x \approx -0.354 \quad \text{or} \quad x \approx -5.646$$

Solving Cubic Equations

Strategy: Find one root by trial (using the Factor Theorem), then reduce to a quadratic.

Example 7: Solve $x^3 - 6x^2 + 11x - 6 = 0$

Step 1: Test simple integer values. Try $x = 1$:

$$1 - 6 + 11 - 6 = 0 \quad \checkmark$$

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

Step 2: Divide $x^3 - 6x^2 + 11x - 6$ by $(x - 1)$ using polynomial long division or synthetic division:

$$x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6)$$

Step 3: Factor the quadratic $x^2 - 5x + 6 = (x - 2)(x - 3)$

Solutions: $x = 1$, $x = 2$, $x = 3$

Verification with Vieta's:

• Sum: $1 + 2 + 3 = 6 = -(-6)/1$ ✓

• Product: $1 \times 2 \times 3 = 6 = -(-6)/1$ ✓

Solving Equations with Variables on Both Sides

Example 8: Solve $4(2x - 1) = 3(x + 5)$

Expand both sides:

$$8x - 4 = 3x + 15$$

Move $x$ terms left, constants right:

$$8x - 3x = 15 + 4$$

$$5x = 19$$

$$x = \frac{19}{5} = 3.8$$

Solving Rational Equations

Example 9: Solve $\frac{3}{x - 2} + \frac{1}{x + 1} = \frac{4}{x^2 - x - 2}$

First, notice that $x^2 - x - 2 = (x - 2)(x + 1)$. So the LCD is $(x - 2)(x + 1)$.

Restrictions: $x \neq 2$ and $x \neq -1$ (these make the denominator zero).

Multiply every term by $(x - 2)(x + 1)$:

$$3(x + 1) + 1(x - 2) = 4$$

$$3x + 3 + x - 2 = 4$$

$$4x + 1 = 4$$

$$4x = 3$$

$$x = \frac{3}{4}$$

Since $\frac{3}{4} \neq 2$ and $\frac{3}{4} \neq -1$, the solution is valid.

Worked Examples: From Basic to Advanced

Here's where we go beyond what most pages offer. Each example targets a specific skill level and a specific mistake pattern we see students make.

Basic Level (Class 7–8 / Grade 7–8)

Example 10: Solve $5(x + 2) = 3(x - 4) + 18$

$$5x + 10 = 3x - 12 + 18$$

$$5x + 10 = 3x + 6$$

$$2x = -4$$

$$x = -2$$

Verification: LHS = $5(-2 + 2) = 5(0) = 0$. RHS = $3(-2 - 4) + 18 = 3(-6) + 18 = -18 + 18 = 0$. ✓

Example 11: If $7x - 5 = 2x + 15$, find the value of $3x + 2$.

First, solve for $x$:

$$5x = 20 \implies x = 4$$

Then: $3(4) + 2 = 14$.

Students frequently solve for $x$ correctly but forget to substitute back into the asked expression. Read the question twice.

Intermediate Level (Class 9–10 / Grade 9–10)

Example 12: Solve $x^2 - 2x - 15 = 0$

Find two numbers multiplying to $-15$ and adding to $-2$: those are $-5$ and $+3$.

$$(x - 5)(x + 3) = 0$$

$x = 5$ or $x = -3$

Example 13: Solve using the quadratic formula: $3x^2 - 2x - 5 = 0$

$$a = 3, \quad b = -2, \quad c = -5$$

$$\Delta = (-2)^2 - 4(3)(-5) = 4 + 60 = 64$$

$$x = \frac{2 \pm \sqrt{64}}{6} = \frac{2 \pm 8}{6}$$

$$x = \frac{10}{6} = \frac{5}{3} \quad \text{or} \quad x = \frac{-6}{6} = -1$$

Example 14 (Word Problem): The sum of a number and its square is 72. Find the number.

Let the number be $x$:

$$x + x^2 = 72$$

$$x^2 + x - 72 = 0$$

Find factors of $-72$ that add to $1$: $9$ and $-8$.

$$(x + 9)(x - 8) = 0$$

$x = -9$ or $x = 8$

Both are mathematically valid. If the problem specifies a positive number, $x = 8$.

Advanced Level (Class 11+ / Competition Prep)

Example 15: Solve $x^3 - 3x^2 - 4x + 12 = 0$

Test $x = 2$: $8 - 12 - 8 + 12 = 0$ ✓

Factor out $(x - 2)$:

$$x^3 - 3x^2 - 4x + 12 = (x - 2)(x^2 - x - 6) = (x - 2)(x - 3)(x + 2)$$

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

Example 16: If $\alpha$ and $\beta$ are roots of $2x^2 - 7x + 3 = 0$, find $\alpha^2 + \beta^2$.

Using Vieta's: $\alpha + \beta = \frac{7}{2}$ and $\alpha \beta = \frac{3}{2}$

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{7}{2}\right)^2 - 2 \cdot \frac{3}{2} = \frac{49}{4} - 3 = \frac{37}{4}$$

Example 17: Solve $x^4 - 13x^2 + 36 = 0$

Substitute $u = x^2$:

$$u^2 - 13u + 36 = 0$$

$$(u - 4)(u - 9) = 0$$

$$u = 4 \implies x^2 = 4 \implies x = \pm 2$$

$$u = 9 \implies x^2 = 9 \implies x = \pm 3$$

Four solutions: $x = -3, -2, 2, 3$

Example 18 (Simultaneous Equations): Solve the system:

$$2x + 3y = 12$$ $$4x - y = 5$$

From the second equation: $y = 4x - 5$

Substitute into the first:

$$2x + 3(4x - 5) = 12$$

$$2x + 12x - 15 = 12$$

$$14x = 27$$

$$x = \frac{27}{14}$$

$$y = 4\left(\frac{27}{14}\right) - 5 = \frac{108}{14} - \frac{70}{14} = \frac{38}{14} = \frac{19}{7}$$

Example 19: Solve $\frac{x+1}{x-1} - \frac{x-1}{x+1} = \frac{5}{6}$

LCD = $6(x - 1)(x + 1)$. Multiply through:

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

Expand the left side using the difference of squares pattern on the $6[\ldots]$ terms:

$$6[(x+1)^2 - (x-1)^2] = 6[(x^2 + 2x + 1) - (x^2 - 2x + 1)] = 6[4x] = 24x$$

Right side: $5(x^2 - 1)$

$$24x = 5x^2 - 5$$

$$5x^2 - 24x - 5 = 0$$

Using the quadratic formula:

$$x = \frac{24 \pm \sqrt{576 + 100}}{10} = \frac{24 \pm \sqrt{676}}{10} = \frac{24 \pm 26}{10}$$

$$x = 5 \quad \text{or} \quad x = -\frac{1}{5}$$

Both are valid (neither makes the original denominators zero).

Common Mistakes Students Make with Algebraic Equations

These aren't hypothetical. They come from tutoring sessions — the same errors we see week after week.

Mistake 1: Sign Errors When Transposing

Wrong: $3x - 8 = 13 \implies 3x = 13 - 8 = 5$

Right: $3x - 8 = 13 \implies 3x = 13 + 8 = 21 \implies x = 7$

The $-8$ on the left becomes $+8$ on the right. Students who write the "move" mechanically without thinking about why the sign flips make this error constantly.

Mistake 2: Distributing to Only Part of a Bracket

Wrong: $3(x + 4) = 3x + 4$

Right: $3(x + 4) = 3x + 12$

The $3$ multiplies everything inside the bracket. Every term. Always.

Mistake 3: Cancelling Incorrectly in Fractions

Wrong: $\frac{x + 3}{3} = x + 1$ (cancelling the 3s)

Right: $\frac{x + 3}{3} = \frac{x}{3} + 1$

You can only cancel a factor of the entire numerator with the denominator — not a term. This is a consequence of how division distributes over addition.

Mistake 4: Forgetting the Negative Root of a Square

Wrong: $x^2 = 16 \implies x = 4$

Right: $x^2 = 16 \implies x = 4$ or $x = -4$

A squared number is always positive, so both $4^2 = 16$ and $(-4)^2 = 16$.

Mistake 5: Dividing by Zero Without Realising It

Wrong approach to $x(x - 3) = 0$: Divide both sides by $x$, getting $x - 3 = 0$, so $x = 3$.

Right approach: Use the zero-product property. $x = 0$ or $x - 3 = 0$, giving $x = 0$ or $x = 3$.

Dividing by $x$ silently assumes $x \neq 0$, which eliminates one valid solution.

Mistake 6: Forgetting to Check Extraneous Solutions

When you multiply both sides of a rational equation by an expression containing the variable, you might introduce solutions that make the original denominator zero. Always substitute back and verify.

Algebraic Equations in Real Life

Algebra isn't a classroom invention. These equation types model real situations:

Linear equations: If a taxi charges ₹50 base fare plus ₹12 per kilometre, your total fare for $d$ km is $F = 50 + 12d$. Need to stay within a ₹350 budget? Solve $50 + 12d = 350$ to find you can travel 25 km.

Quadratic equations: A ball thrown upward from a 20-metre building at 15 m/s follows $h = -5t^2 + 15t + 20$, where $h$ is height and $t$ is time in seconds. When does it hit the ground? Solve $-5t^2 + 15t + 20 = 0$.

Cubic equations: In engineering, cubic equations model the volume of materials, the flow rate of liquids through pipes, and the deflection of beams under load.

How to Approach Algebraic Equations on Exams [A Practical Framework]

1. Identify the type. Look at the highest power. That tells you the equation type and the maximum number of solutions.

2. Count expected roots. A degree-$n$ equation has at most $n$ real roots. If you find fewer, that's fine. If you find more, you've made an error.

3. Pick the right method. Linear → isolate. Quadratic → try factoring first, then formula. Cubic → trial root + factor. Rational → find LCD, multiply, check for extraneous solutions.

4. Verify every solution. Substitute back into the original equation. Not the simplified version — the original. This catches sign errors, extraneous solutions, and arithmetic slips.

5. Read the question again. Did it ask for $x$, or for $2x + 1$? Did it ask for positive values only? Students lose marks by answering a different question than the one asked.

Practice Problems

Test yourself. Solutions follow each problem — try before peeking.

Problem 1: Solve $7x - 3(2x + 5) = 4$ <details> <summary>Solution</summary>

$7x - 6x - 15 = 4 \implies x - 15 = 4 \implies x = 19$

Verify: $7(19) - 3(2 \cdot 19 + 5) = 133 - 3(43) = 133 - 129 = 4$ ✓ </details>

Problem 2: Solve $x^2 + 5x - 14 = 0$ <details> <summary>Solution</summary>

Factors of $-14$ that add to $5$: $7$ and $-2$.

$(x + 7)(x - 2) = 0 \implies x = -7$ or $x = 2$ </details>

Problem 3: Solve $2x^2 - 3x - 9 = 0$ <details> <summary>Solution</summary>

$\Delta = 9 + 72 = 81$

$x = \frac{3 \pm 9}{4}$

$x = 3$ or $x = -\frac{3}{2}$ </details>

Problem 4: Solve $x^3 - 7x + 6 = 0$ <details> <summary>Solution</summary>

Test $x = 1$: $1 - 7 + 6 = 0$ ✓

$x^3 - 7x + 6 = (x - 1)(x^2 + x - 6) = (x - 1)(x + 3)(x - 2)$

Solutions: $x = 1, \ x = -3, \ x = 2$ </details>

Problem 5: If the roots of $x^2 - px + 12 = 0$ are in the ratio $1:3$, find $p$. <details> <summary>Solution</summary>

Let roots be $k$ and $3k$.

Product: $k \cdot 3k = 12 \implies 3k^2 = 12 \implies k^2 = 4 \implies k = \pm 2$

Sum: $k + 3k = 4k = p$

If $k = 2$: $p = 8$. If $k = -2$: $p = -8$. </details>

Algebraic Equations vs Other Mathematical Statements

FAQs on Algebraic Equations

Q1. What is an algebraic equation in simple words?

An algebraic equation is a math sentence that says two things are equal, using at least one variable (like $x$ or $y$).

For example, $x + 5 = 12$ asks: "What number plus 5 equals 12?"

Q2. What is the difference between an algebraic equation and an algebraic expression?

An equation has an equal sign and can be solved.
An expression doesn't have an equal sign and can only be simplified.

Example:

• $2x + 3$ → expression

• $2x + 3 = 9$ → equation

Q3. How many solutions can an algebraic equation have?

• Linear → 1 solution

• Quadratic → at most 2

• Cubic → at most 3

In general, degree $n$ → at most $n$ real solutions.

Q4. What are the basic rules for solving algebraic equations?

Whatever you do to one side, do to the other.

Steps:

• Add / subtract / multiply / divide

• Combine like terms

• Expand brackets

• Verify solution

Q5. What is the easiest way to solve a quadratic equation?

Try factoring first.

If not easy, use quadratic formula for $ax^2 + bx + c = 0$.

Q6. Can an algebraic equation have no solution?

Yes. Example:
$x + 3 = x + 7$ → $3 = 7$ ❌

This is an inconsistent equation.

Q7. Can an algebraic equation have infinite solutions?

Yes. Example:
$2(x + 3) = 2x + 6$ → always true

Called identity.

Q8. How do algebraic equations appear in competitive exams like JEE and Olympiads?

They appear indirectly:

• Equal roots

• Vieta’s formulas

• Constraints

• Geometry & calculus applications