Class 10 quadratic equations: worked examples and practice
A quadratic equation contains a squared variable as its highest power. Learn to recognise one, solve it, and check whether its roots are real.
1. Put the equation in standard form
Write it as ax² + bx + c = 0, where a ≠ 0. The coefficients include their signs. In 2x² − 7x + 3 = 0, a = 2, b = −7 and c = 3.
Move every term to one side before choosing a method. For example, x(x + 3) = 10 becomes x² + 3x − 10 = 0.
2. Solve by factorisation
Worked example: 2x² − 7x + 3 = 0
- Multiply a and c: 2 × 3 = 6.
- Find two numbers whose product is 6 and whose sum is −7: −6 and −1.
- Split the middle term: 2x² − 6x − x + 3 = 0.
- Group and factor: 2x(x − 3) − 1(x − 3) = (2x − 1)(x − 3) = 0.
- Set each factor to zero: x = ½ or x = 3.
Check: substituting 3 gives 18 − 21 + 3 = 0. Substituting ½ gives ½ − 3½ + 3 = 0.
3. Use the quadratic formula
For ax² + bx + c = 0, the roots are x = (−b ± √(b² − 4ac)) / (2a). The entire numerator is divided by 2a.
For x² − 4x + 1 = 0, a = 1, b = −4 and c = 1. Substitution gives x = (4 ± √12) / 2 = 2 ± √3. This method is useful when integer factorisation is not obvious.
4. Read the discriminant
The discriminant is D = b² − 4ac. For real coefficients:
- D > 0: two distinct real roots.
- D = 0: one repeated real root, often described as two equal roots.
- D < 0: no real roots.
For x² + 6x + 9 = 0, D = 36 − 36 = 0. The repeated root is −3, which also follows from (x + 3)² = 0.
Common mistakes
- Losing the minus sign in b: write b = −7 explicitly before substitution.
- Dividing by x and losing a root: x² − 5x = 0 has roots 0 and 5.
- Forgetting the ± sign in the formula.
- Accepting every algebraic root in a word problem: a negative length may not fit the situation.
Try these questions
Solve each on paper before opening the answer.
1. Solve x² − 9x + 20 = 0.
−4 and −5 multiply to 20 and add to −9. So (x − 4)(x − 5) = 0 and x = 4 or 5.
2. How many real roots does 3x² + 2x + 1 = 0 have?
D = 2² − 4 × 3 × 1 = −8. It has no real roots.
3. A rectangle has width x cm, length (x + 2) cm and area 48 cm². Find its dimensions.
x(x + 2) = 48 gives x² + 2x − 48 = (x + 8)(x − 6) = 0. The algebraic roots are −8 and 6. A width must be positive, so the dimensions are 6 cm by 8 cm.
What to revise next
Explain when you would use factorisation and when you would use the formula. Then make a short card for each discriminant case using the flashcard study workflow. Check your textbook for the full chapter and prescribed exercises.
These are original revision questions, not official exam questions or NCERT solutions.
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