Quadratic Equations Class 10 Extra Questions: Quadratic equations for Class 10 Mathematics is one of the most important topics covered in the subject. Students appearing in the board exam must practice the questions from this chapter to score good marks. Solving the quadratic equations Class 10 extra questions will help students boost their exam preparations.

The quadratic equations Class 10 extra questions are formulated as per the NCERT book, and students must practice them to excel in the exam. Experts prepare these additional questions to help students work on their problem-solving skills. We have provided the quadratic equations Class 10 extra questions in this article to help students ace their upcoming board exams.

Extra Questions With Solutions for CBSE Class 10 Quadratic Equations

We have provided all the essential questions with their solutions in this article to help students in their CBSE board exams. Let us solve these extra questions given below and practice well for the exams:

Quadratic Equations Class 10 Extra Questions

Students should refer to these solutions to learn the diverse approaches while solving NCERT Maths Book in-text exercise and questions.

Q.1. What will be the nature of roots of quadratic equation 2x² + 4x – n = 0?

Solution:
D = b² – 4ac
⇒ 4² – 4 x 2 (-7)
⇒ 16 + 56 = 72 > 0
Hence, roots of quadratic equation are real and unequal.

Q.2. If ax² + bx + c = 0 has equal roots, find the value of c.

Solution:
For equal roots D = 0
i.e., b² – 4ac = 0
⇒ b² = 4 ac
⇒ c = ?24?

Q.3. If a and b are the roots of the equation x² + ax – b = 0, then find a and b.

Solution:
Sum of the roots = a + b = – ?? = – a
Product of the roots = ab = ?? = – b
= a + b = – a and ab = -b
⇒ 2a = -b and a = -1
⇒ b = 2 and a = -1

Q.4. Show that x = – 2 is a solution of 3x² + 13x + 14 = 0.

Solution:
Put the value of x in the quadratic equation,
⇒ LHS = 3x² + 13x + 14
⇒ 3(-2)² + 13(-2) + 14
⇒ 12 – 26 + 14 = 0
⇒ RHS Hence, x = -2 is a solution.

Q.5. Find the discriminant of the quadratic equation 4√2x² + 8x + 2√2 = 0).

Solution:
D = 62 – 4ac = (8)² – 4(4√2)(2√2)
⇒ 64 – 64 = 0

Q.6. State whether the equation (x + 1)(x – 2) + x = 0 has two distinct real roots or not. Justify your answer.

Solution:
(x + 1)(x – 2) + x = 0
⇒ x² – x – 2 + x = 0
⇒ x² – 2 = 0
D = b² – 4ac
⇒ (-4(1)(-2) = 8 > 0
∴ Given equation has two distinct real roots.

Q.7. For what value of k, is 3 a root of the equation 2x² + x + k = 0?

Solution:
3 is a root of 2x² + x + k = 0, when
⇒ 2(3)² + 3 + k = 0
⇒ 18+ 3 + k = 0
⇒ k = – 21

Q.8. Find the values of k for which the quadratic equation 9x² – 3kx + k = 0 has equal roots.

Solution:
For equal roots:
D = 0
⇒ b² – 4ac = 0
⇒ (- 3k)² – 4 × 9 × k = 0
⇒ 9k² = 36k
⇒ k = 4

Q.9. Find the value of k for which the equation x² + k(2x + k – 1)+ 2 = 0 has real and equal roots.

Solution:
Given quadratic equation: x² + k(2x + k-1) + 2 = 0)
= x² + 2kx + (k² – k + 2) = 0
For equal roots, b² – 4ac = 0
⇒ 4k² – 4k² + 4k – 8 = 0
⇒ 4k = 8
⇒ k = 2

Q.10. If -5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, then find the value of k.

Solution:
Since – 5 is a root of the equation 2x² + px – 15 = 0
∴ 2(-5)² + p(-5) – 15 = 0
⇒ 50 – 5p – 15 = 0 or 5p = 35 or p = 7
Again p(x² + x) + k = 0 or 7x² + 7x + k = 0 has equal roots
∴ D = 0
i.e., b² – 4ac = 0 or 49- 4 × 7k = 0
⇒ k = 49/28 = 74

Q.11. Write the set of values of k for which the quadratic equation 2x² + kx + 8 = 0 has real roots.

Solution:
For real roots, D ≥ 0
⇒ b² – 4ac ≥ 0
⇒ k² – 4(2)(8) ≥ 0
⇒ k² – 64 ≥ 0
⇒ k² ≥ 64
⇒ k ≤ -8 and k ≥ 8

Q.12. Find the values of p for which the quadratic equation 4x² + px + 3 = 0 has equal roots.

Solution:
For equal roots;
D = 0
⇒ b² – 4ac = 0
⇒ p² – 4 × 4 × 3 = 0
⇒ p² – 48 = 0
⇒ p² = 48
⇒ p = ± √48
⇒ p = 4√3 or -4√3

Q.13. Solve for x: √13x? – 2√3x – 2√3 = 0

Solution:
√3x² – 2√3x – 2√3 = 0
⇒ √3x² – 3√2x + √2x – 2√3 = 0
⇒ √3x(x – √6) + √2(x – √6) = 0
⇒ (√3x + 2)(x – √6) = 0
⇒ √3x + √2 = 0 or x – √6 = 0
⇒ x = −√2/√3 or x = √6

Q.14. If ad ≠ bc, then prove that the equation
(a² + b²) x² + 2(ac + bd)x + (c²+ d²) = 0 has no real roots.

Solution:
The given quadratic equation is (a² + b²⁾x² + 2(ac + bd)x +(c²+ d²) = 0
D = b² – 4ac
= 4(ac + bd)² – 4(a² + b²) (c²+ d²)
= -4(a²d² + b²c²– 2abcd) = – 4(ad – bc)²
Since ad ≠ bc
Therefore D < 0
Hence, the equation has no real roots.

Q.15. Solve for x: √13x² – 2x – 8√3 = 0

Solution:
√3x² – 2x – 8√3 = 0
By mid term splitting
⇒ √3x² – 6x + 4x – 8√3 = 0
⇒ √3x(x – 2/3) + 4 (x – 2/3) = 0
⇒ (x – 2√3)(√3x + 4) = 0
⇒ Either (x – 2√3) = 0 or (√3x + 4) = 0
⇒ x = −4/3√, 2√3

Q.16. Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0

Solution:
(i) We have, 2x² + kx + 3 = 0
Here, a = 2, b = k, c = 3
D = b² – 4ac = k² – 4 × 2 × 3 = k² – 24 For equal roots
D = 0
i.e., k² – 24 = 0
⇒ ķ² = 24
⇒ k = ± √24
⇒ k = + 2√6

(ii) We have, kx(x – 2) + 6 = 0
⇒ kx² – 2kx + 6 = 0
Here, a = k, b = – 2k, c = 6
For equal roots, we have
D = 0
i.e., b² – 4ac = 0
⇒ (-2k)² – 4 × k × 6 = 0
⇒ 4k² – 24k = 0
⇒ 4k (k – 6) = 0
Either 4k = 0 or k – 6 = 0
⇒ k = 0 or k = 6
But k = 6 (because if k = 0 then given equation will not be a quadratic equation)
So, k = 6.

Q.17. If the roots of the quadratic equation (a – b) x² + (b – c) x + (c – a) = 0 are equal, prove that 2a = b + c.

Solution:
Since the equation (a – b)x² + (b – c) x + (c – a) = 0 has equal roots, therefore discriminant
D = (b – c)² – 4(a – b) (c – a) = 0
⇒ b² + c² – 2bc – 4(ac – a² – bc + ab)
⇒ b² + c² – 2bc – 4ac + 4a² + 4bc – 4ab = 0
⇒ 4a² + b² + c² – 4ab + 2bc – 4ac = 0
⇒ (2a)² + (- b)² + (-c)² + 2(2a) (-b) + 2(-b) (-c) + 2(-c) 2a = 0
⇒ (2a – b – c)² = 0
⇒ 2a – b – c = 0
⇒ 2a = b + c. Hence Proved

Q.18. If the equation (1 + m²)x² + 2mcx + c²– a² = 0 has equal roots, show that c² = a² (1 + m²).

Solution:
The given equation is (1 + m²) x² + (2mc) x + (c²– a²) = 0
Here, A = 1 + m², B = 2mc and C = c² – a²
Since the given equation has equal roots, therefore D = 0 = B² – 4AC = 0.
⇒ (2mc)² – 4(1 + m²) (c² – a²) = 0
⇒ 4m²c² – 4(c² – a² + m²c² – m²a²) = 0
⇒ m²c² – c² + a² – m²c² + m²a² = 0. [Dividing throughout by 4]
⇒ – c² + a² (1 + m²) = 0
⇒ c² = a(1 + m²) Hence Proved

Q.19. If the roots of the equation (c- ab)x² – 2(a²– bc)x + b² – ac = 0 in x are equal, then show that either a = 0 or a³ + b³ + c³ = 3abc.

Solution:
For equal roots D = 0
Therefore 4(a² – bc)² – 4(c² – ab) (b² – ac) = 0
⇒ 4[a⁴ + b²c² – 2a²bc – b²c² + ac³ + ab³ – a²bc] = 0.
⇒ a(a³ + b³ + c³ – 3abc) = 0
Either a = 0 or a³ + b³ + c³ = 3abc

Q.20. If the roots of the quadratic equation
(x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0
are equal, then show that a = b = c.

Solution:
Given (x – a) (x – b) + (x – b) (x – c) + (x – 6) (x – a) = 0
⇒ x² – ax – bx + ab + x² – bx – cx + bc + x² – cx – ax + ac = 0
⇒ 3x² – 2(a + b + c)x + ab + bc + ca = 0
Now, for equal roots D = 0
⇒ B² – 4AC = 0
⇒ 4(a + b + c)² – 12(ab + bc + ca) = 0
4a² + 4b² + 4c² + 8ab + 8bc + 8ca – 12ab – 12bc – 12ca = 0
⇒ 2[2a² + 2b² + 2co – 2ab – 2bc – 2ca] = 0
⇒ 2[(a² + b² – 2ab) + (b² + c² – 2bc) + (c² + a² – 2ca)] = 0
⇒ [(a – b)² + (b – c)² + (c – a)²] = 0
⇒ a – b = 0, b – c = 0, c – a = 0
⇒ a = b, b = c, c = a
⇒ a = b = c

Q.21. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects.

Solution:
Let Shefali’s marks in Mathematics be x.
Therefore, Shefali’s marks in English is (30 – x).
Now, according to question,
⇒ (x + 2) (30 – x – 3) = 210
⇒ (x + 2) (27 – x) = 210
⇒ 27x – x² + 54 – 2x = 210
⇒ 25x – x² + 54 – 210 = 0
⇒ 25x – x² – 156 = 0
⇒ -(x² – 25x + 156) = 0
⇒ x² – 25x + 156 = 0
= x² – 13x – 12x + 156 = 0
⇒ x(x – 13) – 12(x – 13) = 0
⇒ (x – 13) (x – 12) = 0
Either x – 13 or x – 12 = 0
∴ x = 13 or x = 12
Therefore, Shefali’s marks in Mathematics = 13
Marks in English = 30 – 13 = 17
or Shefali’s marks in Mathematics = 12
marks in English = 30 – 12 = 18.

Q.22. The sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Solution:
Let x be the length of the side of first square and y be the length of side of the second square.
Then, x² + y² = 468 …(i)
Let x be the length of the side of the bigger square.
4x – 4y = 24
⇒ x – y = 6 or x = y + 6 …(ii)
Putting the value of x in terms of y from equation (ii), in equation (i), we get
(y + 6)² + y² = 468
⇒ y² + 12y + 36 + y² = 468 or 232 + 12y – 432 = 0
⇒ y² + 6y – 216 = 0
⇒ y² + 18y – 12y – 216 = 0
⇒ y(y + 18) – 12(y + 18) = 0
⇒ (y + 18)(y – 12) = 0
Either y + 18 = 0 or y – 12 = 0
⇒ y = -18 or y = 12
But, sides cannot be negative, so y = 12
Therefore, x = 12 + 6 = 18
Hence, sides of two squares are 18 m and 12 m.

Q.23. If twice the area of a smaller square is subtracted from the area of a larger square; the result is 14 cm². However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm². Determine the sides of the two squares.

Solution:
Let the sides of the larger and smaller squares be x and y respectively. Then
x² – 2y² = 14 …..(i)
and 2x² + 3y² = 203 ……(ii)
Operating (ii) -2 × (i), we get
⇒ 2x² + 3y² – (2x² – 4y²) = 203 – 2 × 14
⇒ 2x² + 3y² – 2x² + 4y² = 203 – 28
⇒ 7y² = 175
⇒ y² = 25
⇒ y ± 15
⇒ y = 5 [∵ Side cannot be negative]
By putting the value of y in equation (i), we get
x² – 2 × 5 = 14
⇒ x² – 50 = 14 or x² = 64
∴ x = 8 or x = 8
∴ Sides of the two squares are 8 cm and 5 cm.

Q.24. If Zeba was younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?

Solution:
Let the present age of Zeba be x years
Age before 5 years = (x – 5) years According to given condition,
⇒ (x – 5)² = 5x + 11
⇒ x² + 25 – 10x = 5x + 11
⇒ x² – 10x – 5x + 25 – 11 = 0
⇒ x² – 15x + 14 = 0
⇒ x² – 14x – x + 14 = 0
⇒ x(x – 14) -1(x – 14) = 0
⇒ (x – 1)(x – 14) = 0
⇒ x – 1 = 0 or x – 14 = 0
x = 1 or x = 14
But present age cannot be 1 year.
∴ Present age of Zeba is 14 years.

Q.25. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.

Solution:
Let the natural number be x
According to the question,
⇒  x + 12 = 60?
= x² + 12x – 160 = 0
= x² + 20x – 8x – 160 = 0
= x(x + 20) -8(x + 20) = 0
= (x + 20) (x – 8) = 0
x = -20 (Not possible) or x = 8
Hence, the required natural number is 8.

Q.26. The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.

Solution:
Let the two consecutive multiples of 7 be x and x +7.
x² + (x + 7)² = 637
= x² + x² + 49 + 14x = 637
= 2x² + 14x – 588 = 0
= x² + 7x – 294 = 0
= x² + 21x – 14x – 294 = 0
= x(x + 21) – 14(x + 21) = 0
= x(x + 14) (x + 21) = 0
= x = 14 or x = -21
The multiples are 14 and 21.

Q.27. Find the positive value(s) of k for which both quadratic equations x² + kx + 64 = 0 and x² – 8x + k = 0 will have real roots.

Solution:
(i) For x² + kx + 64 = 0 to have real roots
⇒ k² – 4(1)(64) ≥ 0 i.e., k² – 256 ≥ 0
⇒ k ≥ ± 16

(ii) For x² – 8x + k = 0 to have real roots
⇒ (-8)² – 4(k) ≥ 0 i.e., 64 – 4k ≥ 0
⇒ k ≤ ± 16
For (i) and (ii) to hold simultaneously k = 16

FAQs on Quadratic Equations Class 10 Extra Questions

Q.1. Is quadratic equations an important chapter for CBSE Class 10 students? 
Ans. Yes, quadratic equations is one of the most important chapters students need to learn in Class 10 to ace the board exams.

Q.2. Where can I access the extra questions for Class 10 quadratic equations? 
Ans. Students can find the quadratic equations Class 10 extra questions from Embibe to prepare for the upcoming board exams.

Q.3. Are roots imaginary in Class 10 quadratic equations? 
Ans. The roots in quadratic equations can be real, equal, unequal, or imaginary.

Q.4. What are the ways to solve a quadratic equation? 
Ans. The quadratic equations can be solved using the square roots, factoring, using the quadratic formula, and completing the square.

Q.5. Can a student quickly solve all the quadratic equations Class 10 extra questions? 
Ans. If a student practices the questions well and thoroughly goes through them, they can quickly solve them and get good marks.