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Class 10 Maths Notes

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Quadratic Equations Class 10 Maths Notes

A quadratic equation has at most two roots, and everything in this chapter is about finding those two or deciding how many there are without finding them. Two methods do the finding, factorisation when the coefficients are kind and the quadratic formula when they are not, and one number, the discriminant b² − 4ac, decides how many real roots exist before you start.

Class:10Subject:MathematicsUnit:IICovers:CBSE 2024-25
7 Key Formulas
DWritten byDeep Narayan
Updated
Key Concept Summary

How does the discriminant tell you the nature of the roots of a quadratic equation?

For ax² + bx + c = 0 the discriminant is D = b² − 4ac. If D > 0 the equation has two distinct real roots, if D = 0 it has two equal real roots, and if D < 0 it has no real roots at all. The roots themselves are x = (−b ± √D) / 2a, so a negative discriminant means the square root cannot be taken over the reals and the roots are imaginary.

01

What a Quadratic Equation Is

Definition

Quadratic equation

An equation of the second degree, written in one variable as ax² + bx + c = 0 with real coefficients and a ≠ 0. The condition a ≠ 0 is not decoration: if a = 0 the x² term vanishes and what is left is linear, not quadratic, so the equation would no longer belong to this chapter.

The standard form every quadratic equation must be put into.
  • x² − 5x + 6 = 0 is quadratic with a = 1, b = −5, c = 6.
  • 2x² − 3x − 5 = 0 is quadratic with a = 2, b = −3, c = −5. The a term comes first; there is no x² term anywhere except at the front.
  • 3x − 7 = 0 is linear, not quadratic, because a = 0.
  • 5 = 0 is not an equation in x at all.
  • x² + 1 = 0 is quadratic with a = 1, b = 0, c = 1. A missing middle term means b = 0, and 0 is a perfectly good coefficient.

Situational problems reducible to a quadratic are not in the course

Older books include a set of problems whose given equations contain the unknown in a denominator and have to be cleared before you get a quadratic. That set has been removed from the 2024-25 syllabus. Work only from the day-to-day problems that give you two quantities with a stated sum or a stated product, and write the quadratic directly.
02

Solution by Factorisation

Factorisation is the short method. You split the middle term so that the expression becomes two brackets, and then you set each bracket equal to zero. It works only when the two factors come out as whole numbers or as simple fractions, which is exactly why the board pairs it with the quadratic formula: one method when the numbers are kind, the other when they are not.

  • x² − 5x + 6 = 0. We need two numbers with product +6 and sum −5: they are −2 and −3. So (x − 2)(x − 3) = 0, giving x = 2 or x = 3.
  • x² + 5x + 6 = 0. We need product +6 and sum +5: they are +2 and +3. So (x + 2)(x + 3) = 0, giving x = −2 or x = −3.
  • 2x² − 5x + 3 = 0. We need product +6 and sum −5, and then split the 2: (2x − 3)(x − 1) = 0, since 2x² − 2x − 3x + 3 = 2x² − 5x + 3. So x = 3/2 or x = 1.
  • Check the middle pair: x = 3/2 gives 2 × 9/4 − 5 × 3/2 + 3 = 9/2 − 15/2 + 3 = −3 + 3 = 0. x = 1 gives 2 − 5 + 3 = 0. Both hold.
One quadratic becomes two linear equations, each giving one root.

Finding the split when the numbers do not jump out

Write down the product ac, then list the factor pairs of ac and check which pair adds or subtracts to b. For 2x² − 5x + 3, ac = 6, the pairs are (1, 6), (2, 3) and their negatives, and only (−3, +2) gives −5. Then split the leading coefficient to match: 2x − 3 and x − 1.
03

Solution by the Quadratic Formula

The quadratic formula works for every quadratic equation, kind coefficients or not. Its advantage over factorisation is that it never fails and never needs a clever guess; its cost is that it is arithmetic, so careless sign work loses the mark. Derive it once from completing the square and you will never misremember which way the sign goes.

The quadratic formula, with the discriminant pulled out for convenience.
  • Solve 2x² − 3x − 5 = 0. Here a = 2, b = −3, c = −5, so D = (−3)² − 4 × 2 × (−5) = 9 + 40 = 49.
  • √D = √49 = 7, so x = (3 ± 7) / 4.
  • Taking the plus sign: x = (3 + 7) / 4 = 10/4 = 5/2.
  • Taking the minus sign: x = (3 − 7) / 4 = −4/4 = −1.
  • The two roots are 5/2 and −1, which match the factorisation (2x − 5)(x + 1) = 0 exactly.

The sign of b is the whole difficulty

The formula starts with −b, so when the middle term is −3x you must use −(−3) = +3, and then plus or minus √D gives (3 + 7)/4 and (3 − 7)/4. If you drop the minus sign in front of b you get negative roots that are wrong, and if you forget the minus sign on the first b in a case like x² − 5x + 6 = 0 you will get D = 25 + 24 = 49 instead of 25 − 24 = 1. Write D = b² − 4ac and then read off a, b, c in a separate line.

Two roots means two marks even on a two-mark question

A common short question gives the bigger root only, or the smaller root only, or the sum or product of the roots. Work out both values and then select, and make it clear which is which by writing them as two separate lines.
04

The Discriminant and the Nature of the Roots

The discriminant D = b² − 4ac is the quantity under the square root in the quadratic formula, so it alone decides how many real roots there are before you do any solving. Most of the one-mark and two-mark questions in this chapter are really this table, so learn it as a table and not as a paragraph.

  • D > 0: the equation has two distinct real roots. Example: x² − 5x + 6 = 0 has D = 25 − 24 = 1 > 0, and the roots are 2 and 3.
  • D = 0: the equation has two equal real roots, which is really one root written twice. Example: x² + 4x + 4 = 0 has D = 16 − 16 = 0, and the root is x = −4/2 = −2.
  • D < 0: the equation has no real roots, because the square root of a negative number cannot be taken over the reals. Example: x² + x + 1 = 0 has D = 1 − 4 = −3 < 0.
  • A D = 0 graph touches the x-axis at one point. A D > 0 graph cuts it at two points. A D < 0 graph never meets the x-axis at all.

Two equal roots, not one root

When D = 0, write the roots as two equal values, −b/2a and −b/2a, and say the roots are equal. Do not write a single root and do not say there is no solution. D = 0 means the solution exists and is repeated, which is exactly what happens with x² + 4x + 4 = 0 where x = −2 satisfies 4 − 8 + 4 = 0.

Use D for the nature and the formula for the values

When a question asks whether the roots are real or imaginary, compute D and stop there. When it asks for the roots, then use the formula. Doing both in full when only one is asked wastes time that you may need later in the paper.
05

Zeros and Coefficients of a Quadratic

If the roots of ax² + bx + c = 0 are α and β, then the roots determine the coefficients. This is the same pair of relations you met in Polynomials, and in this chapter it saves you solving at all — several three-mark questions are answered entirely from the sum and the product.

Sum and product of the roots of a quadratic equation.
  • For 2x² − 3x − 5 = 0 with roots 5/2 and −1: the sum is 5/2 − 1 = 3/2, and −b/a = 3/2. The product is −5/2, and c/a = −5/2.
  • The sum of the squares of the roots is (α + β)² − 2αβ. For x² − 5x + 6 = 0 with roots 2 and 3 this gives 5² − 2 × 6 = 25 − 12 = 13, and indeed 2² + 3² = 13.
  • Written in terms of the coefficients, α² + β² = (b² − 2ac)/a². For x² − 5x + 6 that is (25 − 12)/1 = 13, the same answer.
  • The difference of the roots is √D / |a|. For 2x² − 3x − 5 = 0 that is 7/2, and indeed 5/2 − (−1) = 7/2.

The question that needs no solving at all

If the roots of x² − 6x + k are equal, then D = 0, so 36 − 4k = 0 and k = 9. The equation is then x² − 6x + 9 = 0, that is (x − 3)² = 0, and the equal roots are both 3. Put a different polynomial in front of any similar question and the method never changes: compute D, set it to zero, solve for the unknown.
06

Day-to-Day Situational Problems

A situational problem is not a different mathematics. You choose one unknown, translate one sentence into an equation, and the equation that appears is quadratic because the sentence mentions two quantities whose product is fixed. Numbers, rectangular gardens, work and time all fit that pattern.

  • Numbers: the product of two consecutive positive integers is 306. Let them be x and x + 1, so x(x + 1) = 306, which is x² + x − 306 = 0, that is (x + 18)(x − 17) = 0. So x = 17 or x = −18, and only x = 17 is a positive integer. The numbers are 17 and 18, and 17 × 18 = 306.
  • Garden: the length of a rectangular garden is 5 m more than the width and the area is 84 m². Let the width be x, so (x + 5)x = 84, giving x² + 5x − 84 = 0, that is (x + 12)(x − 7) = 0. The width is 7 m and the length is 12 m.
  • Work: A can finish a piece of work in 12 days and B in 18 days. Working together they take x days, so x/12 + x/18 = 1. Multiplying by 36 gives 3x + 2x = 36, so x = 36/5, that is 7 1/5 days.
  • Check the work answer: in 36/5 days A does 36/60 = 3/5 of the work and B does 36/90 = 2/5, and 3/5 + 2/5 = 1, so the work is finished exactly.
  • Speed: a car covers 240 km at a certain speed. Had the speed been 20 km/h more, the journey would have taken 1 hour less. Let the speed be s km/h, so 240/s − 240/(s + 20) = 1, and multiplying by s(s + 20) gives 240(s + 20) − 240s = s(s + 20), so 4800 = s² + 20s, that is (s − 60)(s + 80) = 0.
  • A speed cannot be negative, so s = 60 km/h, and the time taken is 240/60 = 4 hours. Check: at 80 km/h the journey takes 240/80 = 3 hours, exactly 1 hour less.

Reject the impossible root every time

A negative width, a negative number of articles or a negative time is not a solution of the problem even though it solves the equation. State plainly that the negative root is rejected because the quantity cannot be negative, and that line is usually worth a mark of its own.
07

A Complete Five-Mark Problem, Step by Step

Question: a rectangular garden has a length 5 m more than its width and an area of 84 m². Find the dimensions of the garden, and find the cost of fencing it at ₹75 per metre.

  • Let the width be x metres. Then the length is (x + 5) metres.
  • Area gives x(x + 5) = 84, so x² + 5x − 84 = 0. This is the standard form ax² + bx + c = 0 with a = 1, b = 5, c = −84.
  • Factorise. We need a product of −84 and a sum of +5: that is +12 and −7. So (x + 12)(x − 7) = 0, giving x = −12 or x = 7.
  • A width cannot be negative, so reject x = −12 and take x = 7. The width is 7 m and the length is 7 + 5 = 12 m.
  • Perimeter = 2(length + width) = 2(12 + 7) = 2 × 19 = 38 m.
  • Cost of fencing = 38 × ₹75 = ₹2850.
  • Check: 7 × 12 = 84 m², matching the stated area, and 12 − 7 = 5 m, matching the stated difference.
The whole chain: equation, standard form, factorisation, root, perimeter, cost.

What the five marks are for

One mark for the equation with a sensible variable, one for putting it in standard form, one for the factorisation, one for rejecting the negative root and stating the dimensions, and one for the fencing part. Writing every stage on its own line is what makes those five marks visible to the examiner.
08

How the Questions Are Asked

Quadratic Equations closes Unit II Algebra, which carries 20 of the 80 written marks. The questions come in a small number of shapes and the shape is visible from the wording: ask for the nature of the roots and you only need the discriminant; ask for the roots and you need a method; ask about a day-to-day situation and you need a variable chosen carefully.

  • Find the nature of the roots of 2x² − 5x + 3 = 0. (1 to 2 marks, by discriminant alone.)
  • Solve 2x² − 3x − 5 = 0 by the quadratic formula. (3 marks, Section C.)
  • Find the roots of x² + 5x + 6 = 0 by factorisation. (2 marks, Section B.)
  • If the roots of x² + kx + 6 are 2 and 3, find k. (2 marks, Section B.)
  • Show that the equation x² + 4x + 4 = 0 has two equal real roots. (3 marks.)
  • The product of two consecutive positive integers is 306. Find the integers. (5 marks, Section D.)
  • A rectangular garden is 5 m longer than it is wide and has an area of 84 m². Find its dimensions. (5 marks.)

Paper reminders

The Class 10 paper is 80 marks in 3 hours with 38 questions. Q1-18 are one-mark MCQs and Q19-20 are one-mark assertion-reason, Q21-25 carry 2 marks, Q26-31 carry 3 marks, Q32-35 carry 5 marks and Q36-38 are 4-mark case studies, with internal choice in two questions each of Sections B, C and D. Take π = 22/7 unless the question says otherwise, calculators are not allowed, and the remaining 20 marks are internal assessment, 10 for the pen-paper test and multiple assessment, 5 for the portfolio and 5 for the lab practical.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Standard form

The condition a ≠ 0 is what makes it quadratic; without it the equation is linear.

Quadratic formula

Works for every quadratic. Read off a, b and c before substituting.

Discriminant

The quantity under the square root; it alone decides how many real roots exist.

Nature of the roots

The one-mark question in its entirety. Remember that D = 0 means equal roots, not no solution.

Sum and product of the roots

The minus sign belongs to the sum only. Answers many questions without solving.

Sum of the squares of the roots

Expand (α − β)² if you prefer; both routes give the same expression.

Difference of the roots

Useful when a question asks how far apart the two roots are.

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • A quadratic equation must be in the form ax² + bx + c = 0 with a ≠ 0; a missing middle term means b = 0, and that is a legitimate coefficient.
  • The quadratic formula is x = (−b ± √(b² − 4ac)) / 2a. The −b is the step students get wrong, so read off a, b, c on a separate line first.
  • D = b² − 4ac decides the nature of the roots: D > 0 two distinct real roots, D = 0 two equal real roots, D < 0 no real roots.
  • D = 0 means the roots are equal, not absent. Write the single value twice and say the roots are equal.
  • α + β = −b/a and αβ = c/a. Several three-mark questions are answered from these two alone, with no solving at all.
  • For α² + β² use (α + β)² − 2αβ, which becomes (b² − 2ac)/a². For x² − 5x + 6 = 0 it gives 13, matching 2² + 3².
  • In a situational problem, reject the negative root in a sentence of its own — a negative width, count or time is not a solution of the problem.
  • Do not attempt problems whose equations are reducible to a quadratic. They were removed from the 2024-25 course.
  • Unit II Algebra carries 20 of the 80 written marks; the other 20 marks are internal assessment, 10 for the pen-paper test and multiple assessment, 5 for the portfolio and 5 for the lab practical.

FAQ

Frequently asked questions

How do you decide the nature of the roots without solving the equation?

Compute the discriminant D = b² − 4ac. If D > 0 the equation has two distinct real roots, if D = 0 it has two equal real roots, and if D < 0 it has no real roots, because the square root of a negative number is not defined over the reals. For x² − 5x + 6 = 0, D = 25 − 24 = 1 > 0, so the roots are distinct and real, and they turn out to be 2 and 3.

What is the difference between factorisation and the quadratic formula?

Factorisation splits the quadratic into two brackets and needs two numbers whose product is ac and whose sum is b, so it only works neatly when those numbers come out as whole numbers or simple fractions. The quadratic formula x = (−b ± √(b² − 4ac)) / 2a works for every quadratic equation. For 2x² − 5x + 3 = 0, factorisation gives (2x − 3)(x − 1) = 0 and the roots 3/2 and 1, and the formula with D = 25 − 24 = 1 gives the same two roots.

Why does a quadratic equation have at most two roots?

Because it is of degree 2. After substitution or factorisation each variable equation that survives is linear and contributes one root, and a quadratic cannot be split into more than two linear factors. That is also why D = 0 is described as two equal roots: there are two roots in the counting sense, but their common value is −b/2a, as in x² + 4x + 4 = 0 where the repeated root is −2.

How is a word problem turned into a quadratic equation?

Take one unknown for a quantity that the question relates to another, then translate one stated condition into an equation. If the product of two consecutive positive integers is 306, let them be x and x + 1 and write x(x + 1) = 306, which is x² + x − 306 = 0 or (x + 18)(x − 17) = 0. The roots are 17 and −18, and −18 is rejected because the integers are positive, leaving 17 and 18.

What is the relationship between the roots and the coefficients of a quadratic equation?

If the roots are α and β then α + β = −b/a and αβ = c/a. So for 2x² − 3x − 5 = 0 the sum of the roots is 3/2 and the product is −5/2, matching the roots 5/2 and −1. These relations let you answer questions about the sum, the product, the sum of squares or the difference of the roots without ever solving the equation.

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