Mastering Quadratic Equations Class 10 – Smart Learning with FUZY MATH ACADEMY

Mastering Quadratic Equations — Class 10

Smart learning notes and practice from FUZY MATH ACADEMY

Introduction

Quadratic equations are everywhere — in geometry, motion, area problems and many contest questions. This guide gives a clear definition, the key formulas, two main solving methods (factor and formula), worked examples and a big practice set: 30 questions with solutions, arranged from easy to hard. Use the toggles to hide or reveal answers while you practice.

Definition

A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0 where a, b, c are real numbers and a ≠ 0. The highest power of x is 2.

Key formulas

Standard form: ax² + bx + c = 0

Quadratic formula (roots):

x = (-b ± √(b² − 4ac)) / (2a)

Discriminant: Δ = b² − 4ac. It tells us the nature of roots:

• Δ > 0: two distinct real roots
• Δ = 0: one real root (double root)
• Δ < 0: two complex roots

Solution methods

Factor method (when possible)

Rewrite the quadratic as a product of two linear factors: ax² + bx + c = (mx + p)(nx + q) = 0. Then set each factor to zero.

Works best when integer factors exist or after simple manipulation.

Quadratic formula

Always works. Use x = (-b ± √(b² − 4ac)) / (2a). Check the discriminant first to know what to expect.

Useful for non-factorable quadratics or when coefficients are not nice.

Worked examples (step-by-step)

1. Example 1 (Factor method): Solve x² − 5x + 6 = 0.
   Solution

   Try to factor: find two numbers that multiply to 6 and add to −5. Those are −2 and −3. So:

   (x − 2)(x − 3) = 0  →  x = 2 or x = 3

2. Example 2 (Formula method): Solve 2x² + 3x − 5 = 0.
   Solution

   Here a = 2, b = 3, c = −5. Discriminant Δ = 3² − 4·2·(−5) = 9 + 40 = 49.

   x = (-3 ± √49) / (2·2) = (-3 ± 7)/4
   So x = (4)/4 = 1  or  x = (-10)/4 = -5/2

3. Example 3 (When Δ = 0): Solve x² − 6x + 9 = 0.
   Solution

   Δ = (−6)² − 4·1·9 = 36 − 36 = 0. So there is one repeated root:

   x = 6 / 2 = 3  →  x = 3 (double root)

Practice set — 30 questions (toggle answers)

Start with easy ones. Try to solve before opening the answer. Each item contains the question and a collapsible solution.

1. Q1. Solve x² − 7x + 12 = 0.
   Answer & solution

   Factor: (x − 3)(x − 4) = 0 → x = 3, 4.

2. Q2. Solve x² + 5x + 6 = 0.
   Answer & solution

   (x + 2)(x + 3) = 0 → x = −2, −3.

3. Q3. Solve 2x² − x − 3 = 0.
   Answer & solution

   Factor: (2x + 3)(x − 1) = 0 → x = 1, x = −3/2.

4. Q4. Solve 3x² + 7x + 2 = 0.
   Answer & solution

   Factor: (3x + 1)(x + 2) = 0 → x = −1/3, −2.

5. Q5. Solve x² − 4 = 0.
   Answer & solution

   (x − 2)(x + 2) = 0 → x = ±2.

6. Q6. Solve 4x² − 12x + 9 = 0.
   Answer & solution

   Δ = 144 − 144 = 0 → x = 12/(8) = 3/2 (double root). Alternatively (2x − 3)² = 0.

7. Q7. Solve x² + x − 6 = 0.
   Answer & solution

   (x + 3)(x − 2) = 0 → x = 2, −3.

8. Q8. Solve 5x² − 20 = 0.
   Answer & solution

   5(x² − 4) = 0 → x = ±2.

9. Q9. Solve 2x² + 4x + 2 = 0.
   Answer & solution

   Divide by 2: x² + 2x + 1 = 0 → (x + 1)² = 0 → x = −1.

10. Q10. Solve x² − x − 12 = 0.
    Answer & solution

    (x − 4)(x + 3) = 0 → x = 4, −3.

11. Q11. Solve 6x² − x − 2 = 0.
    Answer & solution

    Use formula: Δ = 1 + 48 = 49 → x = (1 ± 7)/12 → x = 8/12 = 2/3 or x = −6/12 = −1/2.

12. Q12. Solve x² + 4x + 5 = 0.
    Answer & solution

    Δ = 16 − 20 = −4 → complex roots x = (−4 ± 2i)/2 = −2 ± i.

13. Q13. Solve 2x² − 9x + 10 = 0.
    Answer & solution

    Factor: (2x − 5)(x − 2) = 0 → x = 5/2, 2.

14. Q14. Solve x² − 3x + 2 = 0.
    Answer & solution

    (x − 1)(x − 2) = 0 → x = 1, 2.

15. Q15. Solve 3x² − 2x − 1 = 0.
    Answer & solution

    Δ = 4 + 12 = 16 → x = (2 ± 4)/6 → x = 1 or x = −1/3.

16. Q16. Solve x² − 2x + 5 = 0.
    Answer & solution

    Δ = 4 − 20 = −16 → x = (2 ± 4i)/2 = 1 ± 2i.

17. Q17. Solve 4x² + 4x + 1 = 0.
    Answer & solution

    This is (2x + 1)² = 0 → x = −1/2 (double root).

18. Q18. Solve x² + 6x + 8 = 0.
    Answer & solution

    (x + 2)(x + 4) = 0 → x = −2, −4.

19. Q19. Solve 7x² − 14x + 7 = 0.
    Answer & solution

    Divide by 7: x² − 2x + 1 = 0 → (x − 1)² = 0 → x = 1.

20. Q20. Solve 2x² + x − 6 = 0.
    Answer & solution

    Δ = 1 + 48 = 49 → x = (−1 ± 7)/4 → x = 6/4 = 3/2 or x = −8/4 = −2.

21. Q21. Solve x² − x + 1 = 0.
    Answer & solution

    Δ = 1 − 4 = −3 → x = (1 ± i√3)/2 (complex).

22. Q22. Solve 9x² − 24x + 16 = 0.
    Answer & solution

    (3x − 4)² = 0 → x = 4/3.

23. Q23. Solve x² + 7x + 10 = 0.
    Answer & solution

    (x + 5)(x + 2) = 0 → x = −5, −2.

24. Q24. Solve 2x² − 5x + 2 = 0.
    Answer & solution

    Δ = 25 − 16 = 9 → x = (5 ± 3)/4 → x = 2 or x = 1/2.

25. Q25. Solve x² − 10x + 21 = 0.
    Answer & solution

    (x − 3)(x − 7) = 0 → x = 3, 7.

26. Q26. Solve 5x² + x − 6 = 0.
    Answer & solution

    Δ = 1 + 120 = 121 → x = (−1 ± 11)/10 → x = 1 or x = −6/5.

27. Q27. Solve x² + 2x − 35 = 0.
    Answer & solution

    (x + 7)(x − 5) = 0 → x = 5, −7.

28. Q28. Solve 6x² + 11x − 35 = 0.
    Answer & solution

    Factor: (2x + 7)(3x − 5) = 0 → x = 5/3, −7/2.

29. Q29. Solve x² − 4x + 2 = 0.
    Answer & solution

    Δ = 16 − 8 = 8 → x = (4 ± √8)/2 = 2 ± √2.

30. Q30. Solve 3x² − x + 4 = 0.
    Answer & solution

    Δ = 1 − 48 = −47 → complex roots x = (1 ± i√47)/6.

20 FAQs (quick answers)

Common questions students ask. Use them as a quick revision.

1. Q: What makes an equation quadratic?
   A: The highest power of x is 2 and coefficient of x² is not zero.
2. Q: What's the discriminant?
   A: Δ = b² − 4ac; it tells the nature of the roots.
3. Q: When can we factor a quadratic?
   A: When you can find two numbers that multiply to ac and add to b (simple integer factors help).
4. Q: When should I use the quadratic formula?
   A: Use it when factoring is hard or impossible with integers; it always works.
5. Q: Can a quadratic have only one root?
   A: Yes — when Δ = 0 the root is repeated.
6. Q: What if Δ < 0?
   A: Roots are complex conjugates (non-real).
7. Q: Are there shortcuts for perfect square trinomials?
   A: Yes: a² ± 2ab + b² = (a ± b)² pattern helps.
8. Q: How do I check my answer?
   A: Substitute the root(s) back into the original equation; it should make the left side zero.
9. Q: Can coefficients be fractions?
   A: Yes — multiply through by the denominator to clear fractions first if that helps.
10. Q: How to solve word problems with quadratics?
    A: Translate the condition to algebraic expressions, form ax² + bx + c = 0 and solve.
11. Q: What is Vieta's formula?
    A: If roots are r1 and r2 for ax² + bx + c = 0, then r1 + r2 = −b/a and r1·r2 = c/a.
12. Q: Can quadratic equations model areas?
    A: Yes, area relations often become quadratic in length variables.
13. Q: How many roots can a quadratic have?
    A: At most two roots (real or complex).
14. Q: Does order of operations matter when using formula?
    A: Yes — compute b² − 4ac before square root and divide by 2a after.
15. Q: Is completing the square useful?
    A: Yes — it's the method behind the quadratic formula and helpful for proofs and derivations.
16. Q: What if coefficients are large?
    A: Use the formula carefully and simplify fractions; calculator helps but understand steps.
17. Q: Any tips to avoid mistakes?
    A: Check signs, simplify before plugging into formula, and verify by substitution.
18. Q: How to handle negative leading coefficient?
    A: You can multiply the whole equation by −1 to make it positive if needed.
19. Q: How long should I practice?
    A: Short daily practice with mixed difficulty is better than one long session.
20. Q: Where to find more NCERT examples?
    A: The NCERT textbook (Class 10, Chapter 4) and its exemplar problems are great places to practice.

NCERT-style word problems (selected)

The following are standard word problems similar to those in NCERT chapter exercises. Attempt them and use the toggles to check answers.

1. W1. The product of two consecutive positive integers is 56. Find the integers.
   Solution

   Let smaller integer be x. Then x(x + 1) = 56 → x² + x − 56 = 0. Δ = 1 + 224 = 225 → x = (−1 ± 15)/2 → positive root x = 7. So integers 7 and 8.

2. W2. A two-digit number is such that the product of its digits is 18 and the number formed by reversing the digits exceeds the original by 9. Find the number.
   Solution

   Let digits be x (tens) and y (units): 10x + y is the number. Given xy = 18 and 10y + x = 10x + y + 9 → 9y − 9x = 9 → y − x = 1.

   So y = x + 1 and x(x + 1) = 18 → x² + x − 18 = 0 → x = 3 or x = −6 (reject). So x = 3, y = 4 → number 34.

3. W3. The area of a rectangle is 48 cm². If its length is increased by 2 cm and width decreased by 2 cm, the area remains the same. Find the original dimensions.
   Solution

   Let width = x, length = 48/x. Given (48/x + 2)(x − 2) = 48. Multiply out and simplify to a quadratic. Equivalent simpler set up: Let dimensions be (x + 2) and (y − 2) with xy = 48; proceed to form quadratic and solve. (Students can reduce to one variable and solve.)

4. W4. From NCERT exemplar: Had Ajita scored 10 more marks in her mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get?
   Solution

   Let actual marks = x. Given 9(x + 10) = x² → x² − 9x − 90 = 0. Δ = 81 + 360 = 441 → x = (9 ± 21)/2 → positive root x = 15. So she scored 15.

(For full NCERT exercise practice see the NCERT textbook Chapter 4 — these examples follow the same style and difficulty.)

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