RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6 PDF Download
RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6 Quadratic Equations has been provided here. Students can refer to these questions before their examinations for better preparation.
Neha Tanna16 Dec, 2024
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RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6: Chapter 8, Exercise 8.6 of RD Sharma Class 10 Maths focuses on solving quadratic equations using the method of completing the square. In this exercise, students learn to transform a given quadratic equation into a perfect square trinomial by adding or subtracting terms.
This method helps in finding the roots of the equation. It also emphasizes the importance of understanding the properties of quadratic equations and their real-life applications.RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6 Overview
RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6 focuses on solving quadratic equations using the factorization method. The key idea is to factorize the quadratic equation of the form ax² + bx + c = 0 and then find the values of x that satisfy the equation. RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6 helps students understand the core concepts of quadratic equations, enhancing their problem-solving skills and laying the foundation for more advanced algebraic techniques. Mastery of quadratic equations is crucial for competitive exams and real-life applications, making it an essential topic in both academic and professional contexts.RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6 PDF
Below, we have provided the RD Sharma Solutions for Class 10 Maths, Chapter 8, Exercise 8.6 on Quadratic Equations. This exercise includes detailed solutions for various types of quadratic equation problems, helping students understand the concepts better. To make learning more convenient, you can download the PDF of the solutions, which will serve as an effective reference for solving exercises and preparing for exams.RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6 PDF
RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6 Quadratic Equations
Below is the RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6 Quadratic Equations -1. Determine the nature of the roots of the following quadratic equations.
Important Notes:
– A quadratic equation is in the form ax 2 +bx +c =0 – To find the nature of roots, first, find determinant “D” – D = b 2 – 4ac – If D > 0, the equation has real and distinct roots. – If D < 0, the equation has no real roots. – If D = 0, the equation has 1 root.(i) 2x 2 -3x + 5 =0
Solution:
Here, a= 2, b= -3, c= 5 D = b 2 – 4ac = (-3) 2 -4(2)(5) = 9 – 40 = -31<0 It’s seen that D<0, and hence, the given equation does not have any real roots.(ii) 2x 2 -6x + 3=0
Solution:
Here, a= 2, b= -6, c= 3 D = (-6) 2 -4(2)(3) = 36 – 24 = 12>0 It’s seen that D>0, and hence, the given equation have real and distinct roots.(iii) (3/5)x 2 – (2/3) + 1 = 0
Solution:
Here, a= 3/5, b= -2/3, c= 1 D = (-2/3) 2 -4(3/5)(1) = 4/9 – 12/5 = -88/45<0 It’s seen that D<0, and hence, the given equation does not have any real roots.(iv) 3x 2 – 4√3x + 4 = 0
Solution:
Here, a= 3, b= – 4 √ 3, c= 4 D = (- 4 √ 3) 2 -4(3)(4) = 48 – 48 = 0 It’s seen that D = 0, and hence, the given equation has only 1 real and equal root.(v) 3x 2 – 2√6x + 2 = 0
Solution:
Here, a= 3, b= – 2 √ 6, c= 2 D = (- 2 √ 6) 2 – 4(3)(2) = 24 – 24 = 0 It’s seen that D = 0, and hence, the given equation has only 1 real and equal root.2. Find the values of k for which the roots are real and equal in each of the following equations.
(i) kx 2 + 4x + 1 = 0
Solution:
The given equation kx 2 + 4x + 1 = 0 is in the form of ax 2 + bx + c = 0 Where a = k, b = 4, c = 1 For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ 4 2 – 4(k)(1) = 0 ⇒ 16 – 4k = 0 ⇒ k = 4 The value of k is 4.(ii) kx 2 – 2√5x + 4 = 0
Solution:
The given equation kx 2 – 2√5x + 4 = 0 is in the form of ax 2 + bx + c = 0 Where a = k, b = – 2√5, c = 4 For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (- 2√5) 2 – 4(k)(4) = 0 ⇒ 20 –16k = 0 ⇒ k = 5/4 The value of k is 5/4.(iii) 3x 2 – 5x + 2k = 0
Solution:
The given equation 3x 2 – 5x + 2k = 0 is in the form of ax 2 + bx + c = 0 Where a = 3, b = -5, c = 2k For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (-5) 2 – 4(3)(2k) = 0 ⇒ 25 – 24k = 0 ⇒ k = 25/24 The value of k is 25/24.(iv) 4x 2 + kx + 9 = 0
Solution:
The given equation 4x 2 + kx + 9 = 0 is in the form of ax 2 + bx + c = 0 Where a = 4, b = k, c = 9 For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ k 2 – 4(4)(9) = 0 ⇒ k 2 – 144 = 0 ⇒ k = ± 12 The value of k is 12 or -12.(v) 2kx 2 – 40x + 25 = 0
Solution:
The given equation 2kx 2 – 40x + 25 = 0 is in the form of ax 2 + bx + c = 0 Where a = 2k, b = -40, c = 25 For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (-40) 2 – 4(2k)(25) = 0 ⇒ 1600 – 200k = 0 ⇒ k = 8 The value of k is 8.(vi) 9x 2 – 24x + k = 0
Solution:
The given equation 9x 2 – 24x + k = 0 is in the form of ax 2 + bx + c = 0 Where a = 9, b = -24, c = k For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (-24) 2 – 4(9)(k) = 0 ⇒ 576 – 36k = 0 ⇒ k = 16 The value of k is 16.(vii) 4x 2 – 3kx + 1 = 0
Solution:
The given equation 4x 2 – 3kx + 1 = 0 is in the form of ax 2 + bx + c = 0 Where a = 4, b = -3k, c = 1 For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (-3k) 2 – 4(4)(1) = 0 ⇒ 9k 2 – 16 = 0 ⇒ k = ± 4/3 The value of k is ± 4/3.(viii) x 2 – 2(5 + 2k)x + 3(7 + 10k) = 0
Solution:
The given equation x 2 – 2(5 + 2k)x + 3(7 + 10k) = 0 is in the form of ax 2 + bx + c = 0 Where a = 1, b = -2(5 + 2k), c = 3(7 + 10k) For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (-2(5 + 2k)) 2 – 4(1)( 3(7 + 10k)) = 0 ⇒ 4(5 + 2k) 2 – 12(7 + 10k) = 0 ⇒ 25 + 4k 2 + 20k – 21 – 30k = 0 ⇒ 4k 2 – 10k + 4 = 0 ⇒ 2k 2 – 5k + 2 = 0 [dividing by 2] Now, solving for k by factorisation, we have ⇒ 2k 2 – 4k – k + 2 = 0 ⇒ 2k(k – 2) – 1(k – 2) = 0 ⇒ (k – 2)(2k – 1) = 0, k = 2 and k = 1/2, So, the value of k can either be 2 or 1/2.(ix) (3k +1)x 2 + 2(k +1)x + k = 0
Solution:
The given equation (3k +1)x 2 + 2(k +1)x + k = 0 is in the form of ax 2 + bx + c = 0 Where a = (3k +1), b = 2(k + 1), c = k For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (2(k + 1)) 2 – 4(3k +1)(k) = 0 ⇒ 4(k +1) 2 – 4(3k 2 + k) = 0 ⇒ (k + 1) 2 – k(3k + 1) = 0 ⇒ 2k 2 – k – 1 = 0 Now, solving for k by factorisation, we have ⇒ 2k 2 – 2k + k – 1 = 0 ⇒ 2k(k – 1) + 1(k – 1) = 0 ⇒ (k – 1)(2k + 1) = 0, k = 1 and k = -1/2, So, the value of k can either be 1 or -1/2.(x) kx 2 + kx + 1 = – 4x 2 – x
Solution:
The given equation kx 2 + kx + 1 = -4x 2 – x This can be rewritten as, (k + 4)x 2 + (k + 1)x + 1 = 0 Now, this in the form of ax 2 + bx + c = 0 Where a = (k +4), b = (k + 1), c = 1 For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (k + 1) 2 – 4(k +4)(1) = 0 ⇒ (k +1) 2 – 4k – 16 = 0 ⇒ k 2 + 2k + 1 – 4k – 16 = 0 ⇒ k 2 – 2k – 15 = 0 Now, solving for k by factorisation, we have ⇒ k 2 – 5k + 3k – 15 = 0 ⇒ k(k – 5) + 3(k – 5) = 0 ⇒ (k + 3)(k – 5) = 0, k = -3 and k = 5, So, the value of k can either be -3 or 5.(xi) (k + 1)x 2 + 2(k + 3)x + (k + 8) = 0
Solution:
The given equation (k +1)x 2 + 2(k +3)x + (k +8) = 0 is in the form of ax 2 + bx + c = 0 Where a = (k +1), b = 2(k + 3), c = (k + 8) For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (2(k + 3)) 2 – 4(k +1)(k + 8) = 0 ⇒ 4(k +3) 2 – 4(k 2 + 9k + 8) = 0 ⇒ (k + 3) 2 – (k 2 + 9k + 8) = 0 ⇒ k 2 +6k + 9 – k 2 – 9k – 8 = 0 ⇒ -3k + 1 = 0 ⇒ k = 1/3 So, the value of k is 1/3.(xii) x 2 – 2kx + 7k – 12 = 0
Solution:
The given equation x 2 – 2kx + 7k – 12 = 0 is in the form of ax 2 + bx + c = 0 Where a = 1, b = -2k, c = 7k – 12 For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (-2k) 2 – 4(1)(7k – 12) = 0 ⇒ 4k 2 – 4(7k – 12) = 0 ⇒ k 2 – 7k + 12 = 0 Now, solving for k by factorisation, we have ⇒ k 2 – 4k – 3k + 12 = 0 ⇒ (k – 4)(k – 3) = 0, k = 4 and k = 3, So, the value of k can either be 4 or 3.(xiii) (k + 1)x 2 – 2(3k + 1)x + 8k + 1 = 0
Solution:
The given equation (k +1)x 2 – 2(3k +1)x + 8k + 1 = 0 is in the form of ax 2 + bx + c = 0 Where a = (k +1), b = -2(3k + 1), c = 8k + 1 For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (-2(3k + 1)) 2 – 4(k +1)(8k + 1) = 0 ⇒ 4(3k +1) 2 – 4(k + 1)(8k + 1) = 0 ⇒ (3k + 1) 2 – (k + 1)(8k + 1) = 0 ⇒ 9k 2 + 6k + 1 – (8k 2 + 9k + 1) = 0 ⇒ 9k 2 + 6k + 1 – 8k 2 – 9k – 1 = 0 ⇒ k 2 – 3k = 0 ⇒ k(k – 3) = 0 Either k = 0 Or, k – 3 = 0 ⇒ k = 3, So, the value of k can either be 0 or 3(xiv) 5x 2 – 4x + 2 + k(4x 2 – 2x + 1) = 0
Solution:
The given equation 5x 2 – 4x + 2 + k(4x 2 – 2x + 1) = 0 This can be rewritten as, x 2 (5 + 4k) – x(4 + 2k) + 2 – k = 0 Now, this is in the form of ax 2 + bx + c = 0 Where a = (4k +5), b = -(2k + 4), c = 2 – k For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (-(2k + 4)) 2 – 4(4k +5)(2 – k) = 0 ⇒ (2k +4) 2 – 4(4k +5)(2 – k) = 0 ⇒ 16 + 4k 2 + 16k – 4(10 – 5k + 8k – 4k 2 ) = 0 ⇒ 16 + 4k 2 + 16k – 40 + 20k – 32k + 16k 2 = 0 ⇒ 20k 2 + 4k – 24 = 0 ⇒ 5k 2 + k – 6 = 0 Now, solving for k by factorisation, we have ⇒ 5k 2 + 6k – 5k – 6 = 0 ⇒ 5k(k – 1) + 6(k – 1) = 0 ⇒ (k – 1)(5k + 6) = 0, k = 1 and k = -6/5, So, the value of k can either be 1 or -6/5.(xv) (4 – k)x 2 + (2k + 4)x + (8k + 1) = 0
Solution:
The given equation (4 – k)x 2 + (2k + 4)x + (8k + 1) = 0 is in the form of ax 2 + bx + c = 0 Where a = (4 – k), b = (2k + 4), c = (8k + 1) For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (2k + 4) 2 – 4(4 – k)(8k + 1) = 0 ⇒ 4k 2 + 16k + 16 – 4(-8k 2 + 32k + 4 – k) = 0 ⇒ 4k 2 + 16k + 16 + 32k 2 – 124k – 16 = 0 ⇒ 36k 2 – 108k = 0 Taking common, ⇒ 9k(k -3 ) = 0 Now, either 9k = 0 ⇒ k = 0 or k – 3 = 0 ⇒ k = 3, So, the value of k can either be 0 or 3.(xvi) (2k + 1)x 2 + 2(k + 3)x + (k +5) = 0
Solution:
The given equation (2k +1)x 2 + 2(k +3)x + (k + 5) = 0 is in the form of ax 2 + bx + c = 0 Where a = (2k +1), b = 2(k + 3), c = (k + 5) For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (2(k + 3)) 2 – 4(2k +1)(k + 5) = 0 ⇒ 4(k +3) 2 – 4(2k 2 + 11k + 5) = 0 ⇒ (k + 3) 2 – (2k 2 + 11k + 5) = 0 [dividing by 4 both sides] ⇒ k 2 + 5k – 4 = 0 Now, solving for k by completing the square, we have ⇒ k 2 + 2 x (5/2) x k + (5/2) 2 = 4 + (5/2) 2 ⇒ (k + 5/2) 2 = 4 + 25/4 = √ 41/4 ⇒ k + (5/2) = ± √ 41/2 ⇒ k = ( √ 41 – 5)/2 or –( √ 41 + 5)/2 So, the value of k can either be ( √ 41 – 5)/2 or –( √ 41 + 5)/2(xvii) 4x 2 – 2(k + 1)x + (k + 4) = 0
Solution:
The given equation 4x 2 – 2(k +1)x + (k + 4)= 0 is in the form of ax 2 + bx + c = 0 Where a = 4, b = -2(k + 1), c = (k + 4) For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (-2(k + 1)) 2 – 4(4)(k + 4) = 0 ⇒ 4(k +1) 2 – 16(k + 4) = 0 ⇒ (k + 1) 2 – 4(k + 4) = 0 ⇒ k 2 – 2k – 15 = 0 Now, solving for k by factorisation, we have ⇒ k 2 – 5k + 3k – 15 = 0 ⇒ k(k – 5) + 3(k – 5) = 0 ⇒ (k – 5)(k + 3) = 0, k = 5 and k = -3, So, the value of k can either be 5 or -3.3. In the following, determine the set of values of k for which the given quadratic equation has real roots:
(i) 2x 2 + 3x + k = 0
Solution:
Given, 2x 2 + 3x + k = 0 It’s of the form of ax 2 + bx + c = 0 Where, a = 2, b = 3, c = k For the given quadratic equation to have real roots D = b 2 – 4ac ≥ 0 D = 9 – 4(2)(k) ≥ 0 ⇒ 9 – 8k ≥ 0 ⇒ k ≤ 9/8 The value of k should not exceed 9/8 to have real roots.(ii) 2x 2 + x + k = 0
Solution:
Given, 2x 2 + x + k = 0 It’s of the form of ax 2 + bx + c = 0 Where, a = 2, b = 1, c = k For the given quadratic equation to have real roots D = b 2 – 4ac ≥ 0 D = 1 2 – 4(2)(k) ≥ 0 ⇒ 1 – 8k ≥ 0 ⇒ k ≤ 1/8 The value of k should not exceed 1/8 to have real roots.(iii) 2x 2 – 5x – k = 0
Solution:
Given, 2x 2 – 5x – k = 0 It’s of the form of ax 2 + bx + c = 0 Where, a = 2, b = -5, c = -k For the given quadratic equation to have real roots D = b 2 – 4ac ≥ 0 D = (-5) 2 – 4(2)(-k) ≥ 0 ⇒ 25 + 8k ≥ 0 ⇒ k ≥ -25/8 The value of k should be lesser than -25/8 to have real roots.(iv) kx 2 + 6x + 1 = 0
Solution:
Given, kx 2 + 6x + 1 = 0 It’s of the form of ax 2 + bx + c = 0 Where, a = k, b = 6, c = 1 For the given quadratic equation to have real roots D = b 2 – 4ac ≥ 0 D = 6 2 – 4(k)(1) ≥ 0 ⇒ 36 – 4k ≥ 0 The given equation will have real roots if, 36 – 4k ≥ 0 36 ≥ 4k 36/4 ≥ k 9 ≥ k ⇒ so, k ≤ 9 The value of k should not exceed 9 to have real roots.(v) 3x 2 + 2x + k = 0
Solution:
Given, 3x 2 + 2x + k = 0 It’s of the form of ax 2 + bx + c = 0 Where, a =3, b = 2, c = k For the given quadratic equation to have real roots D = b 2 – 4ac ≥ 0 D = (2) 2 – 4(3)(k) ≥ 0 ⇒ 4 – 12k ≥ 0 ⇒ 4 ≥ 12k ⇒ k ≤ 1/3 The value of k should not exceed 1/3 to have real roots.4. Find the values of k for which the following equations have real and equal roots.
(i) x 2 – 2(k + 1)x + k 2 = 0
Solution:
Given, x 2 – 2(k + 1)x + k 2 = 0 It’s of the form of ax 2 + bx + c = 0 Where, a =1, b = -2(k + 1), c = k 2 For the given quadratic equation to have real roots D = b 2 – 4ac = 0 D = (-2(k + 1)) 2 – 4(1)(k 2 ) = 0 ⇒ 4k 2 + 8k + 4 – 4k 2 = 0 ⇒ 8k + 4 = 0 ⇒ k = -4/8 ⇒ k = -1/2 The value of k should be -1/2 to have real and equal roots.(ii)k 2 x 2 – 2 (2k – 1)x + 4 = 0
Solution:
Given, k 2 x 2 – 2 (2k – 1)x + 4 = 0 It’s of the form of ax 2 + bx + c = 0 Where, a = k 2 , b = –2 (2k – 1), c = 4 For the given quadratic equation to have real roots D = b 2 – 4ac = 0 D = (-2(2k – 1)) 2 – 4(4)(k 2 ) = 0 ⇒ 4k 2 – 4k + 1 – 4k 2 = 0 [dividing by 4 both sides] ⇒ -4k + 1 = 0 ⇒ k = 1/4 The value of k should be 1/4 to have real and equal roots.(iii) (k + 1)x 2 – 2(k – 1)x + 1 = 0
Solution:
Given, (k + 1)x 2 – 2(k – 1)x + 1 = 0 It’s of the form of ax 2 + bx + c = 0 Where, a = (k + 1), b = -2(k – 1), c = 1 For the given quadratic equation to have real roots D = b 2 – 4ac = 0 D = (-2(k – 1)) 2 – 4(1)(k + 1) = 0 ⇒ 4k 2 – 2k + 1 – k – 1 = 0 [dividing by 4 both sides] ⇒ k 2 – 3k = 0 ⇒ k(k – 3) = 0 ⇒ k = 0 or k = 3 The value of k can be 0 or 3 to have real and equal roots.5. Find the values of k for which the following equations have real roots.
(i) 2x 2 + kx + 3 = 0
Solution:
Given, 2x 2 + kx + 3 = 0 It’s of the form of ax 2 + bx + c = 0 Where, a = 2, b = k, c = 3 For the given quadratic equation to have real roots D = b 2 – 4ac ≥ 0 D = (k) 2 – 4(3)(2) ≥ 0 ⇒ k 2 – 24 ≥ 0 ⇒ k 2 ≥ 24 ⇒ k ≥ 2 √ 6 and k ≤ -2 √ 6 [After taking square root on both sides] The value of k can be represented as (∞, 2 √ 6] U [-2 √ 6, -∞)(ii) kx(x – 2) + 6 = 0
Solution:
Given, kx(x – 2) + 6 = 0 It can be rewritten as, kx 2 – 2kx + 6 = 0 It’s of the form of ax 2 + bx + c = 0 Where, a =k, b = -2k, c = 6 For the given quadratic equation to have real roots D = b 2 – 4ac ≥ 0 D = (-2k) 2 – 4(k)(6) ≥ 0 ⇒ 4k 2 – 24k ≥ 0 ⇒ 4k(k – 6) ≥ 0 ⇒ k ≥ 0 and k ≥ 6 ⇒ k ≥ 6 The value of k should be greater than or equal to 6 to have real roots.(iii) x 2 – 4kx + k = 0
Solution:
Given, x 2 – 4kx + k = 0 It’s of the form of ax 2 + bx + c = 0 Where, a =1, b = -4k, c = k For the given quadratic equation to have real roots D = b 2 – 4ac ≥ 0 D = (-4k) 2 – 4(1)(k) ≥ 0 ⇒ 16k 2 – 4k ≥ 0 ⇒ 4k(4k – 1) ≥ 0 ⇒ k ≥ 0 and k ≥ 1/4 ⇒ k ≥ 1/4 The value of k should be greater than or equal to 1/4 to have real roots.(iv) kx(x – 2√5) + 10 = 0
Solution:
Given, kx(x – 2√5) + 10 = 0 It can be rewritten as, kx 2 – 2√5kx + 10 = 0 It’s of the form of ax 2 + bx + c = 0 Where, a =k, b = -2√5k, c = 10 For the given quadratic equation to have real roots D = b 2 – 4ac ≥ 0 D = (-2√5k) 2 – 4(k)(10) ≥ 0 ⇒ 20k 2 – 40k ≥ 0 ⇒ 20k(k – 2) ≥ 0 ⇒ k ≥ 0 and k ≥ 2 ⇒ k ≥ 2 The value of k should be greater than or equal to 2 to have real roots.(v) kx(x – 3) + 9 = 0
Solution:
Given, kx(x – 3) + 9 = 0 It can be rewritten as, kx 2 – 3kx + 9 = 0 It’s of the form of ax 2 + bx + c = 0 Where, a = k, b = -3k, c = 9 For the given quadratic equation to have real roots D = b 2 – 4ac ≥ 0 D = (-3k) 2 – 4(k)(9) ≥ 0 ⇒ 9k 2 – 36k ≥ 0 ⇒ 9k(k – 4) ≥ 0 ⇒ k ≥ 0 and k ≥ 4 ⇒ k ≥ 4 The value of k should be greater than or equal to 4 to have real roots.(vi) 4x 2 + kx + 3 = 0
Solution:
Given, 4x 2 + kx + 3 = 0 It’s of the form of ax 2 + bx + c = 0 Where, a = 4, b = k, c = 3 For the given quadratic equation to have real roots D = b 2 – 4ac ≥ 0 D = (k) 2 – 4(4)(3) ≥ 0 ⇒ k 2 – 48 ≥ 0 ⇒ k 2 ≥ 48 ⇒ k ≥ 4 √ 3 and k ≤ -4 √ 3 [After taking square root on both sides] The value of k can be represented as (∞, 4 √ 3] U [-4 √ 3, -∞).6. Find the values of k for which the given quadratic equation has real and distinct roots.
(i) kx 2 + 2x + 1 = 0
Solution:
Given, kx 2 + 2x + 1 = 0 It’s of the form of ax 2 + bx + c = 0 Where, a =k, b = 2, c = 1 For the given quadratic equation to have real roots D = b 2 – 4ac > 0 D = (2) 2 – 4(1)(k) > 0 ⇒ 4 – 4k > 0 ⇒ 4k < 4 ⇒ k < 1 The value of k should be lesser than 1 to have real and distinct roots.(ii) kx 2 + 6x + 1 = 0
Solution:
Given, kx 2 + 6x + 1 = 0 It’s of the form of ax 2 + bx + c = 0 Where, a =k, b = 6, c = 1 For the given quadratic equation to have real roots D = b 2 – 4ac > 0 D = (6) 2 – 4(1)(k) > 0 ⇒ 36 – 4k > 0 ⇒ 4k < 36 ⇒ k < 9 The value of k should be lesser than 9 to have real and distinct roots.7. For what value of k, (4 – k)x 2 + (2k + 4)x + (8k + 1) = 0, is a perfect square?
Solution:
Given, (4 – k)x 2 + (2k + 4)x + (8k + 1) = 0 It is in the form of ax 2 + bx + c = 0 Where, a = 4 – k, b = 2k + 4, c = 8k + 1 Calculating the discriminant, D = b 2 – 4ac = (2k + 4) 2 – 4(4 – k)(8k + 1) = 4k 2 + 16 + 4k – 4(32 + 4 – 8k 2 – k) = 4(k 2 + 4 + k – 32 – 4 + 8k 2 + k) = 4(9k 2 – 27k) As the given equation is a perfect square, then D = 0 ⇒ 4(9k 2 – 27k) = 0 ⇒ (9k 2 – 27k) = 0 ⇒ 3k(k – 3) = 0 Thus, 3k = 0 ⇒ k = 0 Or, k-3 = 0 ⇒k = 3 Hence, the value of k should be 0 or 3 for the given to be a perfect square.8. Find the least positive value of k for which the equation x 2 + kx + 4 = 0 has real roots.
Solution:
Given, x 2 + kx + 4 = 0 It’s of the form of ax 2 + bx + c = 0 Where, a =1, b = k, c = 4 For the given quadratic equation to have real roots D = b 2 – 4ac ≥ 0 D = (k) 2 – 4(1)(4) ≥ 0 ⇒ k 2 – 16 ≥ 0 ⇒ k ≥ 4 and k ≤ -4 Considering the least positive value, we have ⇒ k = 4 Thus, the least value of k is 4 for the given equation to have real roots.9. Find the values of k for which the quadratic equation (3k + 1)x 2 + 2(k + 1)x + 1 = 0 has equal roots. Also, find the roots.
Solution:
The given equation (3k +1)x 2 + 2(k +1)x + 1 = 0 is in the form of ax 2 + bx + c = 0 Where a = (3k +1), b = 2(k + 1), c = 1 For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (2(k + 1)) 2 – 4(3k +1)(1) = 0 ⇒ (k +1) 2 – (3k + 1) = 0 [After dividing by 4 both sides] ⇒ k 2 + 2k + 1 – 3k – 1 = 0 ⇒ k 2 – k = 0 ⇒ k(k – 1) = 0 Either k = 0 Or, k – 3 = 0 ⇒ k = 1, So, the value of k can either be 0 or 1 Now, using k = 0 in the given quadratic equation, we get (3(0) + 1)x 2 + 2(0 + 1)x + 1 = 0 x 2 + 2x + 1 = 0 ⇒ (x + 1) 2 = 0 Thus, x = -1 is the root of the given quadratic equation. Next, by using k = 1 in the given quadratic equation, we get (3(1) + 1)x 2 + 2(1 + 1)x + 1 = 0 4x 2 + 4x + 1 = 0 ⇒ (2x + 1) 2 = 0 Thus, 2x = -1 ⇒ x = -1/2 is the root of the given quadratic equation.10. Find the values of p for which the quadratic equation (2p + 1)x 2 – (7p + 2)x + (7p – 3) = 0 has equal roots. Also, find the roots.
Solution:
The given equation (2p + 1)x 2 – (7p + 2)x + (7p – 3) = 0 is in the form of ax 2 + bx + c = 0 Where a = (2p +1), b = -(7p + 2), c = (7p – 3) For the equation to have real and equal roots, the condition is D = b 2 – 4ac = 0 ⇒ (-(7p + 2)) 2 – 4(2p +1)( 7p – 3) = 0 ⇒ (7p + 2) 2 – 4(14p 2 + p – 3) = 0 ⇒ 49p 2 + 28p + 4 – 56p 2 – 4p + 12 = 0 ⇒ -7p 2 + 24p + 16 = 0 Solving for p by factorisation, ⇒ -7p 2 + 28p – 4p + 16 = 0 ⇒ -7p(p – 4) -4(p – 4) = 0 ⇒ (p – 4) (-7p – 4) = 0 Either p – 4 = 0 ⇒ p = 4 Or, 7p + 4 = 0 ⇒ p = -4/7, So, the value of k can either be 4 or -4/7 Now, using k = 4 in the given quadratic equation, we get (2(4) + 1)x 2 – (7(4) + 2)x + (7(4) – 3) = 0 9x 2 – 30x + 25 = 0 ⇒ (3x – 5) 2 = 0 Thus, x = 5/3 is the root of the given quadratic equation. Next, by using k = 1 in the given quadratic equation, we get (2(-4/7 ) + 1)x 2 – (7(-4/7 ) + 2)x + (7(-4/7 ) – 3) = 0 x 2 – 14x + 49 = 0 ⇒ (x – 7) 2 = 0 Thus, x – 7 = 0 ⇒ x = 7 is the root of the given quadratic equation.11. If -5 is a root of the quadratic equation 2x 2 + px -15 = 0 and the quadratic equation p(x 2 + x) + k = 0 has equal roots, find the value of k.
Solution:
Given, -5 is as root of 2x 2 + px – 15 = 0 So, on substituting x = -5, the LHS will become zero and satisfy the equation. ⇒ 2(-5) 2 + p(-5) – 15 = 0 ⇒ 50 – 5p – 15 = 0 ⇒ 35 = 5p ⇒ p = 7 Now, substituting the value of p in the second equation, we have (7)(x 2 + x) + k = 0 ⇒ 7x 2 + 7x + k = 0 It’s given that the above equation has equal roots. Thus the discriminant, D = 0 The equation 7x 2 + 7x + k = 0 is in the form of ax 2 + bx + c = 0 Where a = 7, b = 7, c = k D = b 2 – 4ac ⇒ 7 2 – 4(7)(k) = 0 ⇒ 49 – 28k = 0 ⇒ k = 49/28 = 7/4 Therefore, the value of k is 7/4.12. If 2 is a root of the quadratic equation 3x 2 + px -8 = 0 and the quadratic equation 4x 2 – 2px + k = 0 has equal roots, find the value of k.
Solution:
Given, 2 is as root of 3x 2 + px -8 = 0 So, on substituting x = 2, the LHS will become zero and satisfy the equation. ⇒ 3(2) 2 + p(2) – 8 = 0 ⇒ 12 + 2p – 8 = 0 ⇒ 4 + 2p = 0 ⇒ p = -2 Now, substituting the value of p in the second equation, we have 4x 2 – 2(-2)x + k = 0 ⇒ 4x 2 + 4x + k = 0 It’s given that the above equation has equal roots. Thus the discriminant, D = 0 The equation 4x 2 + 4x + k = 0 is in the form of ax 2 + bx + c = 0 Where a = 4, b = 4, c = k D = b 2 – 4ac ⇒ 4 2 – 4(4)(k) = 0 ⇒ 16 – 16k = 0 [dividing by 16 on both sides] ⇒ k = 1 Therefore, the value of k is 1.Benefits of Using RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6
The RD Sharma Solutions for Class 10 Maths Chapter 8 Exercise 8.6 on Quadratic Equations provides several benefits for students preparing for exams or strengthening their mathematical understanding. Here are the key benefits:1. Clear Understanding of Concepts
RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6 help in providing a step-by-step breakdown of each problem, which ensures that students understand the core concepts of quadratic equations. This is particularly helpful for complex problems that require multiple steps.2. Practice with Diverse Problems
RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6 contains a variety of problems that test different aspects of quadratic equations, such as factoring, solving by the quadratic formula, and applying them in word problems. This provides students with ample practice and helps them build a solid foundation in solving quadratic equations.3. Exam Preparation
The solutions can be extremely beneficial for students preparing for exams like the CBSE Class 10 board exams. The exercise includes problems similar to what students might encounter in the exams, enabling them to test their skills and improve time management.4. Clarification of Mistakes
By referring to the RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6 , students can identify their mistakes and understand the correct approach to solving quadratic equations. This instant feedback loop helps them learn from errors and improve their problem-solving techniques.5. Simplified Explanation
RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6 provide clear and concise explanations of each step involved in solving quadratic equations. This makes it easier for students to grasp the methods used, such as factorization or applying the quadratic formula, and helps them avoid common pitfalls.6. Better Understanding of Various Methods
Quadratic equations can be solved by multiple methods: factoring, completing the square, or using the quadratic formula. RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6 give students exposure to different approaches and help them choose the most appropriate method based on the type of problem.RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.6 FAQs
What are the basics of quadratic equations?
A quadratic equation is an algebraic equation of the second degree in x. The quadratic equation in its standard form is ax2 + bx + c = 0, where a and b are the coefficients, x is the variable, and c is the constant term.
Can a be 0 in a quadratic equation?
b and c can be any numbers including zero. If b or c is zero then these terms will not appear. A quadratic equation takes the form ax2 + bx + c = 0 where a, b and c are numbers. The number a cannot be zero.
What is the range of the quadratic equation?
The range of a quadratic function written in standard form f(x)=a(x−h)2+k with a positive a value is f(x)≥k; the range of a quadratic function written in standard form with a negative a value is f(x)≤k.
Who is the father of the quadratic equation?
Muhammad ibn Musa al-Khwarizmi was an Arab mathematician who wrote a revolutionary book on resolving quadratic equations.
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