Quadratic Equations
CBSE Class 10 · Mathematics · Notes, formulas and practice questions
A quadratic equation is a second-degree polynomial equation in one variable, written as ax² + bx + c = 0. This chapter shows how to find its roots by factorisation and by the quadratic formula, use the discriminant to tell the nature of the roots, and solve word problems that lead to quadratic equations.
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A quadratic equation is a second-degree equation in one variable. Its standard form is ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. A number that makes the left side equal to zero when substituted for x is called a root of the equation. Because the expression is of degree 2, the equation can have at most two distinct roots. This basic idea guides every method in this chapter.
The first method of solution is factorisation. Write ax² + bx + c as the product of two linear factors, and then use the fact that a product is zero only when at least one factor is zero. To factorise, split the middle term into two terms whose coefficients add to b and multiply to ac. This method is quick whenever the quadratic has rational factors.
For a quadratic that is not easy to factorise, the chapter develops the quadratic formula from the method of completing the square. The formula x = (−b ± √(b² − 4ac))/(2a) gives the two roots for any values of a, b and c. The expression b² − 4ac that appears in the formula is given the name discriminant. Its value decides whether the roots are real and distinct, real and equal, or not real at all.
The final part applies these tools to word problems. Statements about consecutive numbers, rectangles and speed are translated into algebra, and simplifying them usually gives ax² + bx + c = 0. Solving the equation is only half the work. A length cannot be negative, a number of objects must be an integer, and a speed given in these questions is positive, so roots that do not fit the situation are rejected. The root that remains should be checked against the original statement.
Key terms
- Quadratic equation
- An equation that can be written in the form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. The coefficient a is the number multiplying the x² term; if a were 0 the equation would become linear, not quadratic.
- Root of a quadratic equation
- A value α is a root only when substituting x = α turns ax² + bx + c into 0. A quadratic equation has at most two roots, and the two roots are equal when the quadratic expression is a perfect square.
- Factorisation method
- Write ax² + bx + c as (mx + n)(px + q) and set each bracket equal to zero. Since a product is zero only if one factor is zero, solving the two linear equations mx + n = 0 and px + q = 0 gives the roots.
- Splitting the middle term
- To factorise ax² + bx + c, choose two numbers whose sum is b and product is ac. Use these two numbers to split bx into two separate terms, then take out the common factor from the four terms.
- Quadratic formula
- The formula x = (−b ± √(b² − 4ac))/(2a) gives the roots of ax² + bx + c = 0. It is obtained by completing the square and works for every quadratic equation that has real roots.
- Discriminant
- The expression D = b² − 4ac that appears under the square root in the quadratic formula. It is calculated separately to decide the nature of the roots without solving the whole equation.
- Nature of roots
- When D > 0 the roots are real and different; when D = 0 the roots are real and equal; and when D < 0 the equation has no real roots. For D = 0 the repeated root is x = −b/(2a).
- Inadmissible root
- In a word problem, an algebraic solution that does not satisfy the conditions of the problem, such as a negative breadth, a negative speed or a fractional number of objects. Such roots are rejected, but only when they do not make sense in the context.
Formula sheet
| What | Formula | Notes |
|---|---|---|
| Standard form of a quadratic equation | ax² + bx + c = 0 | a, b and c are real coefficients with a ≠ 0. Bring every term to one side before identifying a, b and c. |
| Quadratic formula | x = (−b ± √(b² − 4ac))/(2a) | Use for ax² + bx + c = 0 with a ≠ 0. The two values are obtained by taking the plus and minus signs; the roots are real only when b² − 4ac ≥ 0. |
| Discriminant | D = b² − 4ac | D > 0 gives two distinct real roots, D = 0 gives two equal real roots, and D < 0 gives no real roots. |
| Factorised form of roots | (mx + n)(px + q) = 0 ⇒ x = −n/m or x = −q/p | m and p must not be zero. This is the algebraic statement behind the factorisation method. |
Practice questions with answers
1. Define a root of a quadratic equation and verify that x = 3 is a root of x² − 5x + 6 = 0.
A root of ax² + bx + c = 0 is any number α for which aα² + bα + c = 0. Substituting x = 3 gives 3² − 5(3) + 6 = 9 − 15 + 6 = 0, so x = 3 is indeed a root.
2. Solve x² − 5x + 6 = 0 by factorisation.
Split the middle term: x² − 2x − 3x + 6 = 0. Grouping gives x(x − 2) − 3(x − 2) = 0, so (x − 2)(x − 3) = 0. Hence x = 2 or x = 3.
3. Use the quadratic formula to solve x² − 4x − 1 = 0.
Here a = 1, b = −4 and c = −1. The discriminant is (−4)² − 4(1)(−1) = 16 + 4 = 20. Therefore x = (4 ± √20)/2 = (4 ± 2√5)/2 = 2 ± √5.
4. For what value of k does kx² + 6x + 3 = 0 have two equal real roots?
Equal roots require discriminant 0, so 36 − 4(k)(3) = 0. This gives 36 − 12k = 0, hence k = 3. The equation becomes 3x² + 6x + 3 = 0, or 3(x + 1)² = 0.
5. Find the discriminant of 2x² − 3x + 5 = 0 and state the nature of its roots.
D = b² − 4ac = (−3)² − 4(2)(5) = 9 − 40 = −31. Since D < 0, the equation has no real roots.
6. Solve 2x² + x − 6 = 0 by factorisation.
Find two numbers whose sum is 1 and product is 2 × (−6) = −12; they are 4 and −3. Splitting the middle term, 2x² + 4x − 3x − 6 = 2x(x + 2) − 3(x + 2) = (2x − 3)(x + 2). Thus x = 3/2 or x = −2.
7. The product of two consecutive positive integers is 306. Find the integers.
Let the integers be n and n + 1. Then n(n + 1) = 306, so n² + n − 306 = 0. Factorising gives (n + 18)(n − 17) = 0, so n = −18 or n = 17. Since the integers are positive, n = 17, and the integers are 17 and 18.
8. A rectangular park has length 4 m more than its breadth, and its area is 192 m². Find the perimeter of the park.
Let the breadth be x m, so the length is x + 4 m. Then x(x + 4) = 192, giving x² + 4x − 192 = 0 = (x + 16)(x − 12). Reject x = −16, so the breadth is 12 m and the length is 16 m. The perimeter is 2(12 + 16) = 56 m.
9. A train travels 360 km at a uniform speed. If its speed had been 5 km/h more, the journey would have taken 1 hour less. Find the original speed.
Let the original speed be x km/h. The times are 360/x and 360/(x + 5), and 360/x − 360/(x + 5) = 1. Multiplying by x(x + 5) gives 360(x + 5) − 360x = x(x + 5), which simplifies to x² + 5x − 1800 = 0. Factorising, (x + 45)(x − 40) = 0, so x = 40 after rejecting x = −45. The original speed is 40 km/h.
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