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Grade 10/ Maths/ Quadratic Equations
Chapter 4 · NCERT Maths 041

Quadratic Equations

Three ways to find the roots, the one formula that always works (and where it comes from), and the discriminant that tells you the answer before you solve. Every idea on a grapher you can drag.

📈 5 topics⏱ ~45 min📝 20-question quiz
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See every root, live

Drag a, b, c and watch the parabola, its roots (orange dots), the vertex, the axis of symmetry, and the discriminant D update together. This one picture is the whole chapter — no clicking around to find it.

Explore · y = a x² + b x + cdrag a, b, c
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The five ideas in this chapter

A quadratic equation is any equation that can be written in the standard form

a x² + b x + c = 0,   where a ≠ 0

Here a, b, c are real numbers and a ≠ 0 — if a were 0 the x² term would disappear and we'd just have a linear equation. The highest power of x is 2, which is why a quadratic has at most two roots.

A root (also called a solution or zero) is a value of x that makes the equation true. Graphically, the equation y = ax² + bx + c draws a parabola, and its roots are exactly the points where that parabola crosses the x-axis.

Quick relations

For ax² + bx + c = 0: sum of roots = −b/a and product of roots = c/a. Handy for checking your answers.

Watch out: before doing anything, rearrange the equation so one side is 0. "x² = 5x − 6" must become "x² − 5x + 6 = 0" first — otherwise your a, b, c are wrong.

If we can write the quadratic as a product of two linear factors equal to zero, then each factor can be set to zero (because if P·Q = 0 then P = 0 or Q = 0). The trick is splitting the middle term: find two numbers that multiply to a·c and add to b.

Worked example · x² − 5x + 6 = 0
  1. Here a·c = 1·6 = 6, and b = −5. Find two numbers that multiply to 6 and add to −5: that's −2 and −3.
  2. Split the middle term: x² − 2x − 3x + 6 = 0.
  3. Group: x(x − 2) − 3(x − 2) = 0 → (x − 2)(x − 3) = 0.
  4. Set each factor to zero: x − 2 = 0 or x − 3 = 0 → x = 2 or x = 3.
Worked example · 2x² + x − 6 = 0 (when a ≠ 1)
  1. a·c = 2·(−6) = −12, b = 1. Two numbers: 4 and −3 (4·−3 = −12, 4 + (−3) = 1).
  2. 2x² + 4x − 3x − 6 = 0 → 2x(x + 2) − 3(x + 2) = 0 → (2x − 3)(x + 2) = 0.
  3. 2x − 3 = 0 or x + 2 = 0 → x = 3/2 or x = −2.
Common mistake: when a ≠ 1, students multiply to b instead of a·c. Always split using product a·c, sum b.

Factorisation is fast when the numbers are friendly — but many quadratics don't factor neatly. The quadratic formula always works. Here's where it comes from, by completing the square:

Derivation
  1. Start with ax² + bx + c = 0. Divide through by a: x² + (b/a)x + c/a = 0.
  2. Move the constant: x² + (b/a)x = −c/a.
  3. Add (b/2a)² to both sides to complete the square on the left.
  4. The left side becomes a perfect square: (x + b/2a)² = (b² − 4ac) / 4a².
  5. Take the square root of both sides: x + b/2a = ± √(b² − 4ac) / 2a.
  6. Isolate x to get the formula.
x = ( −b ± √(b² − 4ac) ) / 2a
Worked example · 3x² − 5x + 2 = 0
  1. a = 3, b = −5, c = 2 → b² − 4ac = 25 − 24 = 1.
  2. x = (5 ± √1) / 6 = (5 ± 1) / 6.
  3. x = 1 or x = 2/3.

The part under the square root, D = b² − 4ac, is the discriminant. Its sign alone tells you the nature of the roots — no need to finish solving:

  • D > 0 → two distinct real roots. The parabola cuts the x-axis at two points.
  • D = 0 → two equal real roots (a repeated root). The parabola just touches the x-axis.
  • D < 0 → no real roots. The parabola misses the x-axis entirely.

Use the live grapher near the top of the page to watch D, the roots, and the parabola change together — try the "D = 0" and "D < 0" presets to see equal roots and no-real-roots for yourself.

Worked example · nature of 2x² − 4x + 3 = 0
  1. D = (−4)² − 4·2·3 = 16 − 24 = −8.
  2. D < 0 → no real roots (this parabola never touches the x-axis).

Most board questions are word problems. The skill is translating the words into a quadratic, solving it, and then rejecting impossible answers (a length or age can't be negative).

Worked example · consecutive integers

"The product of two consecutive positive integers is 306. Find them."

  1. Let the integers be x and x + 1. Then x(x + 1) = 306.
  2. Expand → x² + x − 306 = 0.
  3. Split −306 = +18·−17 (18 − 17 = 1): (x + 18)(x − 17) = 0.
  4. x = −18 or x = 17. Reject −18 (must be positive) → x = 17.
  5. The integers are 17 and 18. Check: 17 × 18 = 306 ✓.
Method

1) Name the unknown. 2) Form the equation. 3) Bring to standard form. 4) Solve (factorise or formula). 5) Check and reject any root that makes no sense.

Check yourself

Competency quiz

Modelled on CBSE's competency-based pattern — MCQ, assertion–reason and case-study items, the kind that now make up about half your board paper.

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Interactive grapher hand-built for trykarkedekho. Content from the rationalised NCERT Class 10 Maths syllabus (ncert.nic.in).

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