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Quadratic Equations Made Easy for Class 10

March 20, 2026

📘 Quadratic Equations – From Zero to Exam-Ready!

Did you know? Quadratic equations don’t just live in your textbook – they appear in projectile motion, business profit calculations, engineering designs, and even in game physics. Mastering them in Class 10 builds a solid foundation for JEE, NEET, and other competitive exams.

🎯 What Exactly Is a Quadratic Equation?

A quadratic equation in one variable is an equation of the form:

ax2+bx+c=0,a0ax^2 + bx + c = 0,\quad a \ne 0
  • aa, bb, and cc are real numbers (called coefficients).
  • xx is the variable.
  • The highest power of xx is 2, so it’s called “quadratic” (from “quadratus” meaning “square”).

In CBSE Class 10 Maths, you must be able to:

  • Recognise quadratic equations.
  • Convert word problems into quadratic equations.
  • Solve them by:
    • Factorisation
    • Completing the square (helpful for JEE later)
    • Using the quadratic formula

📋 Quick Concept Snapshot

ConceptKey IdeaCBSE/JEE Relevance
Standard formWrite as ax² + bx + c = 0Basic but compulsory
Roots / solutionsValues of x that satisfy the equationUsed in many questions
Factorisation methodSplit middle term to factor the quadraticVery important for Class 10
Quadratic formulaDirect formula for rootsBridge to JEE/NEET level algebra
Discriminant (Δ\Delta)Tells number and nature of rootsFrequently asked in exams
Word problemsConvert situations into equationsHigh-scoring if practiced well

🔍 Recognising Quadratic Equations in Different Forms

Not every quadratic equation will already look like ax2+bx+c=0ax^2 + bx + c = 0. You often need to rearrange.

Example forms:

  • 3x25x+2=03x^2 - 5x + 2 = 0 → already in standard form
  • 2x(x3)=52x(x - 3) = 5 → must expand and bring all terms to one side
  • (x2)(x+4)=0(x - 2)(x + 4) = 0 → factor form, can be expanded or solved directly

First step in MOST questions: bring everything to one side so that the right-hand side is 0.


🧠 Method 1: Solving by Factorisation (Class 10 Favourite)

You’ve probably seen this the most in your NCERT.

Example 1: Basic factorisation

Solve:

x25x+6=0x^2 - 5x + 6 = 0

Step 1: Identify aa, bb, cc.

Here, a=1a = 1, b=5b = -5, c=6c = 6.

Step 2: Find two numbers whose:

  • Product = a×c=1×6=6a \times c = 1 \times 6 = 6
  • Sum = b=5b = -5

Numbers: -2 and -3 (because -2 × -3 = 6 and -2 + -3 = -5)

Step 3: Split the middle term using -2 and -3.

x25x+6=x22x3x+6x^2 - 5x + 6 = x^2 - 2x - 3x + 6

Step 4: Group and factor.

x22x3x+6=x(x2)3(x2)x^2 - 2x - 3x + 6 = x(x - 2) - 3(x - 2) =(x3)(x2)= (x - 3)(x - 2)

Step 5: Apply zero product rule.

If (x3)(x2)=0(x - 3)(x - 2) = 0, then either

  • x3=0x - 3 = 0 so x=3x = 3, or
  • x2=0x - 2 = 0 so x=2x = 2

Roots: 2 and 3.


Example 2: Factorisation when a1a \ne 1

Solve:

6x2+11x10=06x^2 + 11x - 10 = 0

Step 1: Identify aa, bb, cc.

a=6a = 6, b=11b = 11, c=10c = -10

Step 2: Find two numbers whose:

  • Product = a×c=6×(10)=60a \times c = 6 \times (-10) = -60
  • Sum = b=11b = 11

Numbers: 15 and -4 (15 × -4 = -60 and 15 + (-4) = 11)

Step 3: Split the middle term.

6x2+11x10=6x2+15x4x106x^2 + 11x - 10 = 6x^2 + 15x - 4x - 10

Step 4: Group and factor.

6x2+15x4x10=3x(2x+5)2(2x+5)6x^2 + 15x - 4x - 10 = 3x(2x + 5) - 2(2x + 5) =(3x2)(2x+5)= (3x - 2)(2x + 5)

Step 5: Apply zero product rule.

  • 3x2=03x - 2 = 0x=23x = \dfrac{2}{3}
  • 2x+5=02x + 5 = 0x=52x = -\dfrac{5}{2}

So, the roots are 2/32/3 and 5/2-5/2.


🧮 Method 2: Quadratic Formula (Your All-Purpose Weapon)

Sometimes factorisation is difficult or impossible (with irrational roots). Then the quadratic formula is the fastest and most reliable method.

For ax2+bx+c=0ax^2 + bx + c = 0, the roots are given by:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The expression under the square root,

Δ=b24ac\Delta = b^2 - 4ac

is called the discriminant.

Example 3: Using the quadratic formula

Solve:

2x27x+3=02x^2 - 7x + 3 = 0

Step 1: Write aa, bb, cc.

a=2a = 2, b=7b = -7, c=3c = 3

Step 2: Calculate the discriminant.

Δ=b24ac=(7)24×2×3=4924=25\Delta = b^2 - 4ac = (-7)^2 - 4 \times 2 \times 3 = 49 - 24 = 25

Step 3: Apply the quadratic formula.

x=b±Δ2ax = \frac{-b \pm \sqrt{\Delta}}{2a} x=(7)±252×2x = \frac{-(-7) \pm \sqrt{25}}{2 \times 2} x=7±54x = \frac{7 \pm 5}{4}

So we get two roots:

  • x=7+54=124=3x = \dfrac{7 + 5}{4} = \dfrac{12}{4} = 3
  • x=754=24=12x = \dfrac{7 - 5}{4} = \dfrac{2}{4} = \dfrac{1}{2}

Roots are 3 and 1/2.


🔍 Discriminant and Nature of Roots (Very Exam-Friendly)

The discriminant Δ=b24ac\Delta = b^2 - 4ac tells you how many and what kind of solutions the quadratic equation has.

  • If Δ>0\Delta > 0Two distinct real roots
  • If Δ=0\Delta = 0Real and equal roots (repeated root)
  • If Δ<0\Delta < 0No real roots, only complex roots (beyond Class 10, but useful for JEE later)

Example 4: Just checking nature of roots

For x24x+4=0x^2 - 4x + 4 = 0:

Here, a=1a = 1, b=4b = -4, c=4c = 4

Δ=(4)24×1×4=1616=0\Delta = (-4)^2 - 4 \times 1 \times 4 = 16 - 16 = 0

So, the equation has real and equal roots.

(If you solve, you’ll find the root is 2 and 2.)


🎯 Concept Focus Box – “Roots” vs “Zeros” vs “Solutions”

  • Roots of equation: Values of xx that satisfy ax2+bx+c=0ax^2 + bx + c = 0
  • Zeros of quadratic polynomial: Values of xx for which ax2+bx+c=0ax^2 + bx + c = 0 (same idea from polynomial chapter)
  • Solutions: Another word for roots

In Class 10 CBSE Maths, these words are often used interchangeably. In JEE-type questions, they may say “zeros” more often.


💼 Real-Life Connection: Quadratics in Motion and Money

  1. Projectile motion (Physics / NEET relevance):
    Height of a ball thrown upwards is often given by an equation like
    h(t)=5t2+20t+1h(t) = -5t^2 + 20t + 1.
    Setting h(t)=0h(t) = 0 and solving gives the times when the ball hits the ground.

  2. Profit and area problems (Commerce & JEE relevance):

    • Maximum area of a rectangle with fixed perimeter leads to quadratic expressions.
    • Profit functions in business can be quadratic, where roots show break-even points.

These real-life links make quadratic equations powerful beyond just exams.


🧩 Word Problem Example (Highly Scoring in Boards)

Example 5: Number problem

“The product of two consecutive positive integers is 132. Find the integers.”

Step 1: Assume the numbers.

Let the first integer be nn.
Then the next consecutive integer is n+1n + 1.

Step 2: Write equation using the given condition.

Product = 132

n(n+1)=132n(n + 1) = 132 n2+n132=0n^2 + n - 132 = 0

Step 3: Solve the quadratic.

Find two numbers with:

  • Product = 132-132
  • Sum = 1

Numbers: 12 and -11 (12 × -11 = -132, 12 + (-11) = 1)

Split the middle term:

n2+n132=n2+12n11n132n^2 + n - 132 = n^2 + 12n - 11n - 132

Group:

=n(n+12)11(n+12)= n(n + 12) - 11(n + 12) =(n11)(n+12)=0= (n - 11)(n + 12) = 0

So,

  • n11=0n - 11 = 0n=11n = 11
  • n+12=0n + 12 = 0n=12n = -12 (reject as numbers are positive)

So, the integers are 11 and 12.


⚠️ Common Mistakes That Cost Marks

1. Forgetting to rearrange to 0

Students often start applying formula/factorisation directly to equations like:

  • x2+5x=3x^2 + 5x = 3

You must first write:

x2+5x3=0x^2 + 5x - 3 = 0

2. Sign mistakes in discriminant

For b24acb^2 - 4ac, carefully include the sign of bb:

  • If b=7b = -7, then b2=(7)2=49b^2 = (-7)^2 = 49, not -49.

3. Dividing by 2a incorrectly

In the quadratic formula,

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The entire numerator must be divided by 2a2a. Do not do:

  • b/2a±Δ-b/2a \pm \sqrt{\Delta} (this is wrong).

4. Skipping final statement

In CBSE board answers, always end with a clear statement like:

  • “Therefore, the roots of the equation are 2 and 3.”
  • “Hence, the required numbers are 11 and 12.”

This makes your solution neat and exam-ready.


📝 Exam Strategy Corner: How to Score 5/5 on Quadratic Equation Questions

  1. For 1-mark questions (MCQs / very short):

    • Quickly use discriminant to check nature of roots.
    • Memorise special forms like (xa)2=0(x - a)^2 = 0 gives equal roots.
  2. For 2-mark questions:

    • Use factorisation method where possible (faster and cleaner).
    • Avoid long calculations; choose smart method.
  3. For 3–4 mark questions:

    • Show all steps clearly.
    • Mention aa, bb, cc before using the quadratic formula.
    • Box or underline the final roots.
  4. For word problems:

    • Define variables clearly: “Let the required number be x”.
    • Write equation step with a short reason: “According to the condition…”
    • Don’t forget to reject meaningless roots (like negative length, negative time).

🧾 Mini Revision Sheet – Quadratic Equations (Class 10) ✅

  • Standard form: ax2+bx+c=0ax^2 + bx + c = 0, a0a \ne 0
  • Roots: Values of xx that satisfy the equation
  • Factorisation method:
    • Find numbers with product acac and sum bb
    • Split middle term, group, factor
  • Quadratic formula:
x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
  • Discriminant:
Δ=b24ac\Delta = b^2 - 4ac
  • Nature of roots:

    • Δ>0\Delta > 0 → two distinct real roots
    • Δ=0\Delta = 0 → real and equal roots
    • Δ<0\Delta < 0 → no real roots (complex roots)
  • Always:

    • Rearrange to get 0 on one side
    • Check calculations carefully
    • Present final answer clearly

🧠 Did You Know? (For Curious Minds and Future Aspirants)

  • In coordinate geometry, every quadratic equation corresponds to a parabola.
  • The roots of the quadratic are the x-intercepts of this parabola.
  • In JEE-level maths, you’ll study how the sign of the discriminant tells you whether the parabola cuts the x-axis at two points, touches it, or does not touch it at all.

Understanding this in Class 10 makes advanced math much easier later!


🎓 Ready to Practise “Quadratic Equations”?

Consistent practice is the secret to mastering this chapter for CBSE boards and building a strong base for JEE and NEET maths.

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