Class 10Maths · AlgebraFull chapter

Quadratic Equations

The whole chapter in one place — read the notes, move the coefficients to watch the parabola and its roots, then test yourself. Clear notes, the quadratic formula and discriminant, an interactive grapher, practice and worked NCERT solutions.

In this chapter
Standard Form of a Quadratic EquationSolving by Factorisation (Splitting the Middle Term)The Quadratic FormulaThe Discriminant and the Nature of RootsWord Problems Using Quadratic Equations

Standard Form of a Quadratic Equation

Quick answer A quadratic equation is a degree-2 polynomial equation that can be written as ax² + bx + c = 0 with a ≠ 0.

An equation is called a quadratic equation when its highest power of the variable is 2. In its standard form we write it as ax2 + bx + c = 0, where a, b, c are real numbers and, most importantly, a ≠ 0. If a were 0, the x2 term would vanish and the equation would only be linear.

Here a is the coefficient of x2, b is the coefficient of x, and c is the constant term. A real number α is called a root (or zero) of the equation if it satisfies it, that is, if aα2 + bα + c = 0. A quadratic equation has at most two roots.

Many equations are not given in standard form and must be rearranged first. For example, take x(x + 4) = 12. Expanding gives x2 + 4x = 12, and bringing every term to one side gives x2 + 4x − 12 = 0. Now we can read off a = 1, b = 4, c = −12.

To verify a root, substitute it back. Is x = 2 a root of x2 − 3x + 2 = 0? Put x = 2: (2)2 − 3(2) + 2 = 4 − 6 + 2 = 0. Since the left side equals 0, x = 2 is indeed a root.

Standard form ax² + bx + c = 0, a ≠ 0 a, b, c are real numbers; a is the coefficient of x².
Root condition aα² + bα + c = 0 α is a root (zero) if it satisfies the equation.
Number of roots At most 2 roots A degree-2 equation cannot have more than two roots.
Remember
  • Standard form is ax² + bx + c = 0 with a, b, c real and a ≠ 0.
  • The condition a ≠ 0 is what makes the equation quadratic, not linear.
  • A quadratic equation has at most two roots (zeroes).
  • Rearrange every term to one side before identifying a, b and c.
  • A number is a root only if substituting it makes the equation equal to 0.

Solving by Factorisation (Splitting the Middle Term)

Quick answer Split the middle term bx into two parts whose product is a×c and whose sum is b, factorise, then use the zero-product rule.

Factorisation turns a quadratic into a product of two linear factors. Then we use the zero-product rule: if a product of two factors is 0, at least one of them must be 0. So from (px + q)(rx + s) = 0 we get px + q = 0 or rx + s = 0.

To factorise ax2 + bx + c, we split the middle term bx into two terms whose:

  • product equals a × c, and
  • sum equals b.

Worked example: Solve 6x2 − x − 2 = 0.

  1. Here a × c = 6 × (−2) = −12, and b = −1.
  2. Find two numbers with product −12 and sum −1. These are −4 and +3 (since −4 × 3 = −12 and −4 + 3 = −1).
  3. Split the middle term: 6x2 − 4x + 3x − 2 = 0.
  4. Group: 2x(3x − 2) + 1(3x − 2) = 0, so (3x − 2)(2x + 1) = 0.
  5. Set each factor to zero: 3x − 2 = 0 gives x = 2/3; 2x + 1 = 0 gives x = −1/2.

So the roots are x = 2/3 and x = −1/2. You can check the first: 6(2/3)2 − (2/3) − 2 = 6(4/9) − 2/3 − 2 = 8/3 − 2/3 − 2 = 2 − 2 = 0. Correct.

Splitting rule find p, q with p × q = a×c and p + q = b Then write bx = px + qx and factorise by grouping.
Zero-product rule If A × B = 0, then A = 0 or B = 0 Each linear factor set to 0 gives one root.
Example result 6x² − x − 2 = 0 ⇒ x = 2/3 or x = −1/2 From (3x − 2)(2x + 1) = 0.
Remember
  • Zero-product rule: if AB = 0 then A = 0 or B = 0.
  • Split bx into two terms with product a×c and sum b.
  • Group in pairs and take out the common binomial factor.
  • Set each linear factor equal to zero to get the two roots.
  • Always verify a root by substituting it back into the equation.
Watch the parabola respond. Open the grapher — move a, b and c and watch the roots, vertex and discriminant change live. Open the grapher

The Quadratic Formula

Quick answer When real roots exist, they are given by x = (−b ± √(b² − 4ac)) / 2a, obtained by completing the square.

Some quadratics are hard to factorise. For every quadratic we can instead use the quadratic formula, which is obtained by completing the square on ax2 + bx + c = 0. The result is:

x = ( −b ± √(b2 − 4ac) ) / 2a

The two roots come from taking the + sign and the − sign in turn. The formula gives real roots only when the quantity under the root, b2 − 4ac, is not negative.

Worked example: Solve 2x2 − 7x + 3 = 0.

  1. Identify a = 2, b = −7, c = 3.
  2. Compute b2 − 4ac = (−7)2 − 4(2)(3) = 49 − 24 = 25.
  3. √25 = 5, and 2a = 4.
  4. x = (−(−7) ± 5) / 4 = (7 ± 5) / 4.
  5. Taking +: x = 12/4 = 3. Taking −: x = 2/4 = 1/2.

So the roots are x = 3 and x = 1/2. Check x = 3: 2(9) − 7(3) + 3 = 18 − 21 + 3 = 0. Correct.

Quadratic formula x = ( −b ± √(b² − 4ac) ) / 2a Valid when b² − 4ac ≥ 0.
The two roots x = (−b + √(b² − 4ac))/2a and x = (−b − √(b² − 4ac))/2a From the + and − signs respectively.
Example result 2x² − 7x + 3 = 0 ⇒ x = 3 or x = 1/2 Here b² − 4ac = 25.
Remember
  • The quadratic formula solves any quadratic equation.
  • It is derived by the method of completing the square.
  • Read off a, b, c carefully, keeping their signs.
  • The ± sign gives the two roots; real roots need b² − 4ac ≥ 0.
  • Simplify the surd and reduce the fraction to get the final roots.

The Discriminant and the Nature of Roots

Quick answer The discriminant D = b² − 4ac decides the nature of the roots: two distinct real, two equal real, or no real roots.

Inside the quadratic formula sits the expression b2 − 4ac. It is so important that it has its own name: the discriminant, written D = b2 − 4ac. Its sign alone tells us the nature of the roots without solving fully:

  • D > 0: two distinct real roots.
  • D = 0: two equal (coincident) real roots, each equal to −b/2a.
  • D < 0: no real roots (the square root of a negative number is not real).

Example 1: For x2 + 4x + 4 = 0, D = (4)2 − 4(1)(4) = 16 − 16 = 0. So the roots are real and equal, each x = −b/2a = −4/2 = −2.

Example 2: For 2x2 + 3x + 4 = 0, D = (3)2 − 4(2)(4) = 9 − 32 = −23 < 0, so it has no real roots.

Example 3 (finding an unknown): For what value of k does 2x2 + kx + 3 = 0 have equal roots? Equal roots need D = 0, so k2 − 4(2)(3) = 0, giving k2 = 24 and k = ± 2√6.

Discriminant D = b² − 4ac The quantity under the root in the quadratic formula.
Two distinct real roots b² − 4ac > 0 The two roots are unequal.
Two equal real roots b² − 4ac = 0, root = −b / 2a The roots coincide.
No real roots b² − 4ac < 0 No real value of x satisfies the equation.
Remember
  • Discriminant D = b² − 4ac decides the nature of the roots.
  • D > 0 gives two distinct real roots.
  • D = 0 gives two equal real roots, each −b/2a.
  • D < 0 means there are no real roots.
  • For equal roots of an unknown-coefficient equation, set D = 0 and solve.

Word Problems Using Quadratic Equations

Quick answer Translate the situation into a quadratic equation, solve it, and reject any root that does not fit the real-life condition.

Many real situations — numbers, ages, areas, speeds and times — lead to quadratic equations. The method is always the same:

  1. Let the unknown quantity be a variable, say x.
  2. Use the given conditions to form a quadratic equation in x.
  3. Solve it by factorisation or the quadratic formula.
  4. Reject any root that is not admissible (for example a negative length or a fractional number of people).

Worked example (consecutive integers): The product of two consecutive positive integers is 306. Find the integers.

  1. Let the smaller integer be x, so the next is x + 1.
  2. Their product: x(x + 1) = 306, which gives x2 + x − 306 = 0.
  3. Discriminant: D = (1)2 − 4(1)(−306) = 1 + 1224 = 1225, and √1225 = 35.
  4. x = (−1 ± 35) / 2, giving x = 17 or x = −18.
  5. The integers must be positive, so we reject x = −18 and take x = 17.

The two consecutive integers are 17 and 18 (and 17 × 18 = 306, which checks out).

Worked example (area): A rectangular plot has area 528 m2 and its length is one more than twice its breadth. Let the breadth be x metres; then the length is (2x + 1) metres. So x(2x + 1) = 528, i.e. 2x2 + x − 528 = 0. Here D = 1 + 4(2)(528) = 4225 and √4225 = 65, giving x = (−1 + 65)/4 = 16 (rejecting the negative root). So the breadth is 16 m and the length is 33 m.

Consecutive integers x(x + 1) = 306 ⇒ x² + x − 306 = 0 ⇒ x = 17 The integers are 17 and 18; reject x = −18.
Rectangle (area) x(2x + 1) = 528 ⇒ 2x² + x − 528 = 0 ⇒ x = 16 Breadth 16 m, length 33 m; reject the negative root.
Method form quadratic → solve → reject inadmissible root Answer must satisfy the real-life condition of the problem.
Remember
  • Assign a variable to the unknown, then form a quadratic from the conditions.
  • Solve by factorisation or the quadratic formula.
  • Reject roots that are physically impossible (negative length, negative count, etc.).
  • Always state the answer in the words of the problem and verify it.
  • Common types: numbers, ages, areas/dimensions, and speed–time problems.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

ax² + bx + c = 0, a ≠ 0
Standard form
aα² + bα + c = 0
Root condition
At most 2 roots
Number of roots
find p, q with p × q = a×c and p + q = b
Splitting rule
If A × B = 0, then A = 0 or B = 0
Zero-product rule
6x² − x − 2 = 0 ⇒ x = 2/3 or x = −1/2
Example result
x = ( −b ± √(b² − 4ac) ) / 2a
Quadratic formula
x = (−b + √(b² − 4ac))/2a and x = (−b − √(b² − 4ac))/2a
The two roots
2x² − 7x + 3 = 0 ⇒ x = 3 or x = 1/2
Example result
D = b² − 4ac
Discriminant
b² − 4ac > 0
Two distinct real roots
b² − 4ac = 0, root = −b / 2a
Two equal real roots
b² − 4ac < 0
No real roots
x(x + 1) = 306 ⇒ x² + x − 306 = 0 ⇒ x = 17
Consecutive integers
x(2x + 1) = 528 ⇒ 2x² + x − 528 = 0 ⇒ x = 16
Rectangle (area)
form quadratic → solve → reject inadmissible root
Method

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Standard form easy

The standard form of a quadratic equation in the variable x is:

Q2 Standard form easy

Which of the following is NOT a quadratic equation?

Q3 Roots by factorisation easy

The roots of the quadratic equation x² − 5x + 6 = 0 are:

Q4 Roots by factorisation medium

The roots of 6x² − x − 2 = 0 are:

Q5 Discriminant easy

The discriminant of the quadratic equation 2x² − 4x + 3 = 0 is:

Q6 Nature of roots easy

The nature of the roots of the equation x² − 4x + 4 = 0 is:

Q7 Nature of roots medium

The values of k for which 2x² + kx + 3 = 0 has two equal real roots are:

Q8 Roots by factorisation medium

If one root of x² + px + 12 = 0 is 4, then the other root is:

Q9 Quadratic formula medium

Using the quadratic formula, the roots of 3x² − 5x + 2 = 0 are:

Q10 Word problems hard

The sum of the squares of two consecutive positive integers is 365. The integers are:

Q11 Roots by factorisation hard

The quadratic equation whose roots are −3 and 5 is:

Q12 Nature of roots hard

The values of k for which x² + 2(k + 2)x + 9k = 0 has equal roots are:

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Find the roots of the quadratic equation 2x² − 5x + 3 = 0 by factorisation.Roots by factorisation

Given: 2x2 − 5x + 3 = 0

Splitting the middle term: product of a and c = 2 × 3 = 6, and we need two numbers with product 6 and sum −5, namely −2 and −3.

  1. 2x2 − 2x − 3x + 3 = 0
  2. 2x(x − 1) − 3(x − 1) = 0
  3. (x − 1)(2x − 3) = 0

So x − 1 = 0 or 2x − 3 = 0.

Roots: x = 1 and x = 3/2.

2 Find the roots of x² − 3x − 10 = 0 by factorisation.Roots by factorisation

Given: x2 − 3x − 10 = 0

We need two numbers with product −10 and sum −3, namely −5 and +2.

  1. x2 − 5x + 2x − 10 = 0
  2. x(x − 5) + 2(x − 5) = 0
  3. (x − 5)(x + 2) = 0

So x − 5 = 0 or x + 2 = 0.

Roots: x = 5 and x = −2.

3 Find the roots of 2x² + x − 4 = 0 using the quadratic formula.Quadratic formula

Given: 2x2 + x − 4 = 0, so a = 2, b = 1, c = −4.

Discriminant: D = b2 − 4ac = (1)2 − 4(2)(−4) = 1 + 32 = 33.

Since D > 0, the roots are real and distinct:

  1. x = (−b ± √D) / (2a) = (−1 ± √33) / 4

Roots: x = (−1 + √33)/4 and x = (−1 − √33)/4.

4 Find the discriminant of 3x² − 2x + 1/3 = 0 and hence find the nature of its roots. Also find the roots if they are real.Discriminant and nature of roots

Given: 3x2 − 2x + 1/3 = 0, so a = 3, b = −2, c = 1/3.

Discriminant: D = b2 − 4ac = (−2)2 − 4(3)(1/3) = 4 − 4 = 0.

Since D = 0, the roots are real and equal.

  1. Equal root = −b / (2a) = 2 / (2 × 3) = 1/3

Roots: x = 1/3 and x = 1/3.

5 Find the value of k for which the quadratic equation kx(x − 2) + 6 = 0 has two equal roots.Nature of roots

Given: kx(x − 2) + 6 = 0, i.e. kx2 − 2kx + 6 = 0, so a = k, b = −2k, c = 6.

For equal roots, D = 0:

  1. b2 − 4ac = 0
  2. (−2k)2 − 4(k)(6) = 0
  3. 4k2 − 24k = 0
  4. 4k(k − 6) = 0

So k = 0 or k = 6. But k = 0 makes the equation non-quadratic, so it is rejected.

Answer: k = 6.

6 Find two numbers whose sum is 27 and whose product is 182.Word problems

Given: Sum of the two numbers = 27 and product = 182.

Let one number be x, then the other is (27 − x).

  1. x(27 − x) = 182
  2. 27x − x2 = 182
  3. x2 − 27x + 182 = 0
  4. x2 − 13x − 14x + 182 = 0
  5. x(x − 13) − 14(x − 13) = 0
  6. (x − 13)(x − 14) = 0

So x = 13 or x = 14, and the other number is 14 or 13 respectively.

Answer: The two numbers are 13 and 14.

Previous-year board questions 4

Q1 Find the value of k for which the quadratic equation 9x² + 6kx + 4 = 0 has equal roots. CBSE 2019 2 marks

Given: 9x2 + 6kx + 4 = 0, so a = 9, b = 6k, c = 4.

For equal roots, discriminant D = 0:

  1. b2 − 4ac = 0
  2. (6k)2 − 4(9)(4) = 0
  3. 36k2 − 144 = 0
  4. k2 = 4

Answer: k = 2 or k = −2, i.e. k = ± 2.

Q2 A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train. CBSE 2018 3 marks

Given: Distance = 360 km. Let the speed be x km/h.

Time at speed x = 360/x hours; time at speed (x + 5) = 360/(x + 5) hours. The faster trip takes 1 hour less:

  1. 360/x − 360/(x + 5) = 1
  2. 360(x + 5) − 360x = x(x + 5)
  3. 1800 = x2 + 5x
  4. x2 + 5x − 1800 = 0

D = 52 − 4(1)(−1800) = 25 + 7200 = 7225, and √7225 = 85.

  1. x = (−5 ± 85)/2

x = 40 or x = −45. Speed cannot be negative, so x = 40.

Answer: Speed of the train = 40 km/h.

Q3 The sum of the reciprocals of a person's ages (in years) 3 years ago and 5 years from now is 1/3. Find the person's present age. CBSE 2023 3 marks

Given: Let the present age be x years. Age 3 years ago = (x − 3), age 5 years from now = (x + 5).

  1. 1/(x − 3) + 1/(x + 5) = 1/3
  2. [(x + 5) + (x − 3)] / [(x − 3)(x + 5)] = 1/3
  3. (2x + 2) / (x2 + 2x − 15) = 1/3
  4. 3(2x + 2) = x2 + 2x − 15
  5. 6x + 6 = x2 + 2x − 15
  6. x2 − 4x − 21 = 0
  7. (x − 7)(x + 3) = 0

So x = 7 or x = −3. Age cannot be negative, so x = 7.

Answer: Present age = 7 years.

Q4 Two water taps together can fill a tank in 9 3/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. CBSE 2020 4 marks

Given: Together the taps fill the tank in 9 3/8 = 75/8 hours. Let the smaller tap take x hours; the larger tap takes (x − 10) hours.

Work done in 1 hour: 1/x + 1/(x − 10) = 8/75.

  1. [(x − 10) + x] / [x(x − 10)] = 8/75
  2. (2x − 10) / (x2 − 10x) = 8/75
  3. 75(2x − 10) = 8(x2 − 10x)
  4. 150x − 750 = 8x2 − 80x
  5. 8x2 − 230x + 750 = 0
  6. 4x2 − 115x + 375 = 0 (dividing by 2)

D = (−115)2 − 4(4)(375) = 13225 − 6000 = 7225, and √7225 = 85.

  1. x = (115 ± 85)/8

x = 25 or x = 30/8 = 3.75. If x = 3.75, then (x − 10) is negative, which is impossible, so x = 25.

Answer: The smaller tap takes 25 hours and the larger tap takes 25 − 10 = 15 hours.

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