In this chapter, we provide RD Sharma Class 10 Ex 8.6 Solutions Chapter 8 Quadratic Equations for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 8.6 Solutions Chapter 8 Quadratic Equations pdf, Now you will get step by step solution to each question.
| Textbook | NCERT |
| Class | Class 10 |
| Subject | Maths |
| Chapter | Chapter 8 |
| Chapter Name | Quadratic Equations |
| Exercise | 8.6 |
| Category | RD Sharma Solutions |
RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations Ex 8.6 Download PDF
Chapter 8: Quadratic Equations Exercise – 8.6
Question: 1
Determine the nature of the roots of the following quadratic equations.
(i) 2x2 – 3x + 5 = 0
(ii) 2x2 – 6x + 3 = 0
(iii) For what value of k (4 – k)x2 + (2k + 4)x + (8k + 1) = 0 is a perfect square.
(iv) Find the least positive value of k for which the equation x2 + kx + 4 = 0 has real roots.
(v) Find the value of k for which the given quadratic equation has real roots and distinct roots.
Kx2 + 2x + 1 = 0
(vi) Kx2 + 6x + 1 = 0
(vii) x2 – kx + 9 = 0
Solution:
The given quadratic equation is in the form of ax2 + bx + c = 0
So a = 2, b = – 3, c = 5
We know, determinant (D) = b2 – 4ac = (-3)2 – 4(2)(5) = 9 – 40 = – 31 < 0
Since D < 0, the determinant of the equation is negative, so the expression does not having any real roots.
(ii) 2x2 – 6x + 3 = 0
The given quadratic equation is in the form of ax2 + bx + c = 0
So a = 2, b = -6, c = 3
We know, determinant (D) = b2 – 4ac = (- 6)2 – 4(2)(3) = 36 – 24 = 12 < 0
Since D > 0, the determinant of the equation is positive, so the expression does having any real and distinct roots
(iii) For what value of k (4 – k)x2 + (2k + 4)x + (8k + 1) = 0 is a perfect square.
The given equation is (4 – k)x2 + (2k + 4)x + (8k + 1) = 0
Here, a = 4 – k, b = 2k + 4, c = 8k + 1
The discriminate (D) = b2 – 4ac = (2k+4)2 – 4(4 – k)(8k + 1)
= (4k2 + 16 + 16k) – 4(32k + 4 – 8k2 – k)
= 4(k2 + 8k2 + 4k – 31k + 4-4)
= 4(9k2 – 27k)
D = 4(9k2 – 27k)
The given equation is a perfect square D = 0
4(9k2 – 27k) = 0
9k2 – 27k = 0
Taking out common of 3 from both sides and cross multiplying = k2 – 3k = 0
= K (k – 3) = 0
Either k = 0 Or k = 3
The value of k is to be 0 or 3 in order to be a perfect square.
(iv)Find the least positive value of k for which the equation x2 + kx + 4 = 0 has real roots.
The given equation is x2 + kx + 4 = 0 has real roots Here, a = 1, b = k, c = 4
The discriminate (D) = b2 – 4ac = 0 = k2 – 16 = 0 = k = 4, k = – 4
The least positive value of k = 4 for the given equation to have real roots.
(v) Find the value of k for which the given quadratic equation has real roots and distinct roots.
Kx2 + 2x + 1 = 0
The given equation is Kx2 + 2x + 1 = 0
Here, a = k, b = 2, c = 1
The discriminate (D) = b2 – 4ac = 0
= 4 – 4k = 0
= 4k = 4
K = 1
The value of k = 1 for which the quadratic equation is having real and equal roots.
(vi) Kx2 + 6x + 1 = 0
The given equation is Kx2 + 6x + 1 = 0
Here, a = k, b = 6, c = 1
The discriminate (D) = b2 – 4ac = 0
= 36 – 4k = 0
= 4k = 36
K = 9
The value of k = 9 for which the quadratic equation is having real and equal roots.
(vii) x2 – kx + 9 = 0
The given equation is x2 – kx + 9 = 0
Here, a = 1, b = – k, c = 9
Given that the equation is having real and distinct roots.
Hence, the discriminate (D) = b2 – 4ac = 0
= k2 – 4(1)(9) = 0
= k2 – 36 = 0
= K = – 6 and k = 6
The value of k lies between -6 and 6 respectively to have the real and distinct roots.
Question: 2
Find the value of k (i) Kx2 + 4x + 1 = 0.
(ii) 
(iii) 3x2 – 5x + 2k = 0
(iv) 4x2 + kx + 9 = 0
(v) 2kx2 – 40x + 25 = 0
(vi) 9x2 – 24x + k = 0
(vii) 4x2– 3kx + 1 = 0
(viii) x2 – 2(5 + 2k)x + 3(7 + 10k) = 0
(ix) (3k +1)x2+ 2(k +1)x + k = 0
(x) Kx2 + kx + 1 = – 4x2 – x
(xi) (k + 1)x2 + 2(k + 3)x + k + 8 = 0
(xii) x2 – 2kx + 7k – 12 = 0
(xiii) (k + 1)x2 – 2(3k + 1)x + 8k + 1 = 0
(xiv) 5x2 – 4x + 2 + k(4x2 – 2x + 1) = 0
(xv) (4 – k)x2 + (2k + 4)x + (8k + 1) = 0
(xvi) (2k + 1)x2 + 2(k + 3)x + (k +5 ) = 0
(xvii) 4x2 – 2(k + 1)x + (k + 4) = 0
Solution:
The given equation Kx2 + 4x + 1 = 0 is in the form of ax2 + bx + c = 0
Where a = k, b = 4, c = 1
Given that, the equation has real and equal roots D = b2 – 4ac = 0
= 42 – 4(k)(1) = 0
= 16 – 4k = 0
= k = 4
The value of k is 4
(ii)
The given equationis in the form of ax2 + bx + c = 0 where a= k, b = – 2√5, c = 4.
Given that, the equation has real and equal roots D = b2– 4ac = 0
20 – 16k = 0
K = 5 /4
The value of k is k = 5/4
(iii) 3x2 – 5x + 2k = 0
The given equation 3x2 – 5x + 2k = 0 is in the form of ax2 + bx + c = 0 where a = 3, b = – 5, c = 2k
Given that, the equation has real and equal roots D = b2 – 4ac = 0
= (- 5)2 – 4(3)(2k) = 0
= 25 – 24k = 0
K = 25/24
The value of the k is k = 25/24
(iv) 4x2 + kx + 9 = 0
The given equation 4x2 + kx + 9 = 0 is in the form of ax2 + bx + c = 0 where a = 4, b = k, c = 9
Given that, the equation has real and equal roots D = b2 – 4ac = 0
= k2 – 4(4)(9) = 0
= k2 – 144 = 0
= k = 12
The value of k is 12
(v) 2kx2 – 40x + 25 = 0
The given equation 2kx2 – 40x + 25 = 0 is in the form of ax2 + bx + c = 0 where a = 2k, b = – 40, c = 25
Given that, the equation has real and equal roots D = b2 – 4ac = 0
(-40)2 – 4(2k)(25) = 0
1600 – 200k = 0
k = 8
The value of k is 8
(vi) 9x2 – 24x + k = 0
The given equation 9x2 – 24x + k = 0 is in the form of ax2 + bx + c = 0 where a = 9, b = – 24, c = k
Given that, the equation has real and equal roots D = b2 – 4ac = 0
( – 24)2 – 4(9)(k) = 0
576 – 36k = 0
k = 16
The value of k is 16
(vii) 4x2– 3kx + 1 = 0
The given equation 4x2 – 3kx + 1 = 0 is in the form of ax2 + bx + c = 0 where a = 4, b = – 3k, c = 1
Given that, the equation has real and equal roots D = b2 – 4ac = 0
= (-3k)2 – 4(4)(1) = 0
= 9k2 – 16 = 0
K = 4/3
The value of k is 4/3
(viii) x2 – 2(5 + 2k)x + 3(7 + 10k) = 0
The given equation X2 – 2(5 + 2k)x + 3(7 + 10k) = 0 is in the form of ax2 + bx + c = 0 where a = 1, b = +2(52k), c = 3(7 + 10k)
Given that, the nature of the roots of the equation are real and equal roots D = b2 – 4ac = 0
= (+2(52k))2 – 4(1)(3(7 + 10k)) = 0
= 4(5 + 2k)2 – 12(7 + 10k) = 0
= 25 + 4k2 + 20k – 21 – 30k = 0
= 4k2 – 10k + 4 = 0
Simplifying the above equation.
We get, = 2k2 – 5k + 2 = 0
= 2k2 – 4k – k + 2 = 0
= 2k(k – 2) – 1(k – 2) = 0
= (k – 2)(2k – 1) = 0, K = 2 and k = 1/2 The value of k can either be 2 or 1/2
(ix) (3k +1)x2+ 2(k +1)x + k = 0
The given equation (3k + 1)x2 + 2(k + 1)x + k = 0 is in the form of ax2 + bx + c = 0 where a = 3k + 1, b = +2(k + 1), c = (k)
Given that, the nature of the roots of the equation are real and equal roots D = b2 – 4ac = 0
= [2(k + 1)]2 – 4(3k + 1)(k) = 0
= (k + 1)2 – k(3k + 1) = 0
= -2k2 + k + 1 = 0
This equation can also be written as 2k2 – k – 1 = 0
The value of k can be obtained by k
The value of k are 1 and (-1)/2 respectively.
(x) Kx2 + kx + 1 = -4x2 – x
Bringing all the x components on one side we get, x2(4 + k) + x(k + 1) + 1 = 0
The given equation Kx2 + kx + 1 = -4x2 – x is in the form of ax2 + bx + c = 0 where a = 4 + k,b = +k + 1, c = 1 Given that, the nature of the roots of the equation are real and equal roots D= b2– 4ac = 0
= (k+1)2– 4(4 + k)(1) = 0
= k2– 2k – 10 = 0
The equation is also in the form ax2 + bx + c = 0
The value of k is obtained by a = 1, b = -2, c = – 15
Putting the respective values in the above formula we will obtain the value of k
The value of k are 5 and -3 for different given quadratic equation.
(xi) (k + 1)x2 + 2(k + 3)x + k + 8 = 0
The given equation (k + 1) x2 + 2(k + 3)x + k + 8 = 0 is in the form of ax2 + bx + c = 0 where a = k + 1,b = 2(k + 3), c = k + 8
Given the nature of the roots of the equation are real and equal. D = b2 – 4ac = 0
= [2(k + 30]2 – 4(k + 1)(k + 8) = 0
= 4(k + 3)2 – 4(k + 1)(k + 8) = 0
Taking out 4 as common from the LHS of the equation and dividing the same on the RHS = (k + 3)2-(k + 1)(k + 8) = 0
= k2 + 9 + 6k – (k2 + 9k + 18) = 0
Cancelling out the like terms on the LHS side = 9 + 6k – 9k – 8 = 0
= – 3k + 1 = 0
= 3k = 1
K = 1/3
The value of k of the given equation is k =1/3
(xii) x2 – 2kx + 7k – 12 = 0
The given equation is x2 – 2kx + 7k – 12 = 0
The given equation is in the form of ax2 + bx + c = 0 where a = 1, b = – 2k, c = 7k – 12
Given the nature of the roots of the equation are real and equal. D = b2 – 4ac = 0
= (2k)2 – 4(1)(7k – 12) = 0
= 4k2 – 28k + 48 = 0
= k2 – 7k + 12 = 0
The value of k can be obtained by
Here a = 1, b = – 7k, c = 12
By calculating the value of k is
The value of k for the given equation is 4 and 3 respectively.
(xiii) (k + 1)x2 – 2(3k + 1)x + 8k + 1 = 0
The given equation is (k + 1)x2 – 2(3k + 1)x + 8k + 1 = 0
The given equation is in the form of ax2 + bx + c = 0 where a = k + 1, b = – 2(k + 1), c = 8k + 1
Given the nature of the roots of the equation are real and equal. D = b2– 4ac = 0
= (-2(k + 1))2 – 4(k + 1)(8k + 1) = 0
= 4(3k + 1)2 – 4(k + 1)(8k + 1) = 0
Taking out 4 as common from the LHS of the equation and dividing the same on the
RHS = (3k + 1)2 – (k + 1)(8k + 1) = 0
= 9k2 + 6k + 1 – (8k2 + 9k + 1) = 0
= 9k2 + 6k + 1 – 8k2 – 9k – 1 = 0
= k2 – 3k = 0
= k(k – 3) = 0
Either k = 0 Or, k – 3 = 0 = k = 3
The value of k for the given equation is 0 and 3 respectively.
(xiv) 5x2 – 4x + 2 + k(4x2 – 2x + 1) = 0
The given equation 5x2 – 4x + 2 + k(4x2 – 2x + 1) = 0 can be written as x2(5 + 4k) – x(4 + 2k) + 2 – k = 0
The given equation is in the form of ax2 + bx + c = 0 where a = 5 + 4k, b = -(4 + 2k), c = 2 – k
Given the nature of the roots of the equation are real and equal.
D = b2 – 4ac = 0
= [-(4 + 2k)]2 – 4(5 + 4k)(2 – k) = 0
= 16 + 4k2 + 16 – 4(10 – 5k + 8k – 4k2] = 0
= 16 + 4k2 + 16 – 40 + 20k – 32k + 16k2 = 0
= 20k2 – 4k – 24 = 0
Taking out 4 as common from the LHS of the equation and dividing the same on the RHS = 5k2 – k – 6 = 0 The value of k can be obtained by equation
The value of k for the given equation are k = 6/5 and−1 respectively.
(xv) (4 – k)x2 + (2k + 4)x + (8k + 1) = 0
The given equation is (4 – k)x2 + (2k + 4)x + (8k + 1) = 0
The given equation is in the form of ax2 + bx + c = 0 where a = 4 – k, b = (2k + 4), c = 8k + 1
Given the nature of the roots of the equation are real and equal.
D = b2 – 4ac = 0
= (2k + 4)2 – 4(4 – k)(8k + 1) = 0
= 4k2 + 16k + 16 – 4(-8k2 + 32k + 4 – k) = 0
= 4k2 + 16k + 16 + 32k2 – 124k – 16 = 0
Cancelling out the like and opposite terms.
We get, = 36k2 – 108k = 0
Taking out 4 as common from the LHS of the equation and dividing the same on the RHS = 9k2 – 27k = 0
= 9k(k -3 ) = 0
Either 9k = 0 K = 0 Or, k – 3 = 0 K = 3
The value of k for the given equation is 0 and 3 respectively.
(xvi) (2k + 1)x2 + 2(k + 3)x + (k +5 )= 0
The given equation is (2k + 1)x2 + 2(k + 3)x + (k + 5) = 0
The given equation is in the form of ax2 + bx + c = 0 where a = 2k + 1, b = 2(k + 3), c = k + 5
Given the nature of the roots of the equation are real and equal.
D = b2 – 4ac = 0
= [2(k + 3)]2 – 4(2k + 1)(k + 5) = 0
Taking out 4 as common from the LHS of the equation and dividing the same on the
RHS = [(k + 3)]2 – (2k + 1)(k + 5) = 0
= K2 + 9 + 6k – (2k2 + 11k + 5) = 0
= – k2 – 5k + 4 = 0
= k2 + 5k – 4 = 0
The value of k can be obtained by k = 6/5 and − 1 respectively.
Here a = 1, b = 5, c = – 4
The value of k for the given equation is
(xvii) 4x2 – 2(k + 1)x + (k + 4) = 0
The given equation is 4x2 – 2(k + 1)x + (k + 4) = 0
The given equation is in the form of ax2 + bx + c = 0 where a = 4, b = -2(k + 1), c = k + 4
Given the nature of the roots of the equation are real and equal.
D = b2 – 4ac = 0
= [-2(k + 1)]2 – 4(4)(k + 4) = 0
Taking out 4 as common from the LHS of the equation and dividing the same on the
RHS = (k + 1)2 – 4(k + 4) = 0
= k2 + 1 + 2k – 4k – 16 = 0
= k2 – 2k – 15 = 0
The value of k can be obtained by k = 6/5 and − 1 respectively.
Here a = 1, b = -2, c = -15
The value of k for the given equation is
Question: 3
In the following, determine the set of values of k for which the given quadratic equation has real roots:
(i) 2x2 + 3x + k = 0
(ii) 2x2 + kx + 3 = 0
(iii) 2x2 – 5x – k = 0
(iv) Kx2 + 6x + 1= 0
(v) x2 – kx + 9 = 0
Solution:
(i) 2x2 + 3x + k = 0
The given equation is 2x2 + 3x +k = 0
The given quadratic equation has equal and real roots D = b2– 4ac = 0
The given equation is in the form of ax2 + bx + c = 0 so, a = 2, b = 3, c = k = 9 – 4(2)(k) = 0
= 9 – 8k = 0
= k ≤ 98
The value of k does not exceed k ≤ 98 to have a real root.
(ii) 2x2 + kx + 3 = 0
The given equation is 2x2 + kx + 3 = 0
The given quadratic equation has equal and real roots D = b2 – 4ac = 0
The given equation is in the form of ax2 + bx + c = 0 so, a = 2, b = k, c = 3 = k2 – 4(2)(3) = 0
= k2 – 24 = 0
The value of k should not exceed
in order to obtain real roots.
(iii) 2x2 – 5x – k = 0
The given equation is 2x2 – 5x – k = 0
The given quadratic equation has equal and real roots D = b2 – 4ac = 0
The given equation is in the form of ax2 + bx + c = 0 so, a = 2, b = – 5, c = – k = 25 – 4(2)(- k) = 0
= 25 – 8k = 0
= k ≤ 25/8
The value of k should not exceed k ≤ 25/8
(iv) Kx2 + 6x + 1= 0
The given equation is Kx2 + 6x + 1 = 0
The given quadratic equation has equal and real roots D = b2 – 4ac = 0
The given equation is in the form of ax2 + bx + c = 0 so, a = k, b = 6, c = 1 = 36 – 4(k)(1) = 0
= 36 – 4k = 0
= k = 9
The value of k for the given equation is k = 9
(v) x2 – kx + 9 = 0
The given equation is X2 – kx + 9 = 0
The given quadratic equation has equal and real roots D = b2– 4ac = 0
The given equation is in the form of ax2 + bx + c = 0 so, a = 1, b = -k, c = 9 = k2 – 4(1)(-9) = 0
= k2 – 36 = 0
= k2 = 36 k ≥ √36 K = 6 and k = – 6
The value of k should in between K = 6 and k = – 6 in order to maintain real roots.
Question: 4
Determine the nature of the roots of the following quadratic equations.
(i) 2x2 – 3x + 5 = 0
(ii) 2x2 – 6x + 3 = 0
(iii) For what value of k (4 – k)x2 + (2k + 4)x + (8k + 1) = 0 is a perfect square
(iv) Find the least positive value of k for which the equation x2 + kx + 4 = 0 has real roots.
(v) Find the value of k for which the given quadratic equation has real roots and distinct roots.
Kx2 + 2x + 1 = 0
(vi) Kx2 + 6x + 1 = 0
(vii) x2 – kx + 9 = 0
Solution:
(i) 2x2 – 3x + 5 = 0
The given quadratic equation is in the form of ax2 + bx + c = 0
So a = 2, b = -3, c = 5
We know, determinant (D) = b2 – 4ac = (-3)2 – 4(2)(5)
= 9 – 40 = – 31 < 0
Since D < 0, the determinant of the equation is negative, so the expression does not having any real roots.
(ii) 2x2 – 6x + 3 = 0
The given quadratic equation is in the form of ax2 + bx + c = 0
So a = 2, b = -6, c = 3
We know, determinant (D) = b2 – 4ac = (-6)2 – 4(2)(3) = 36 – 24 = 12 < 0
Since D > 0, the determinant of the equation is positive, so the expression does having any real and distinct roots.
(iii) For what value of k (4 – k)x2 + (2k + 4)x + (8k + 1) = 0 is a perfect square
The given equation is (4 – k)x2 + (2k + 4)x + (8k + 1) = 0
Here, a = 4 – k, b = 2k + 4, c = 8k + 1
The discriminate (D) = b2 – 4ac
= (2k + 4)2 – 4(4 – k)(8k + 1)
= (4k2 + 16 + 16k) – 4(32k + 4 – 8k2 – k)
= 4(k2 + 8k2 + 4k – 31k + 4 – 4)
= 4(9k2 – 27k)
= 4(9k2 – 27k)
The given equation is a perfect square D = 0
4(9k2 – 27k) = 0
9k2 – 27k = 0
Taking out common of 3 from both sides and cross multiplying K2 – 3k = 0 K (k – 3) = 0
Either k = 0 Or k = 3
The value of k is to be 0 or 3 in order to be a perfect square.
(iv) Find the least positive value of k for which the equation x2 + kx + 4 = 0 has real roots.
The given equation is x2 + kx + 4 = 0 has real roots Here, a = 1, b = k, c = 4
The discriminate (D) = b2 – 4ac = 0
= k2 – 16 = 0
= k = 4, k = – 4
The least positive value of k = 4 for the given equation to have real roots.
(v) Find the value of k for which the given quadratic equation has real roots and distinct roots.
Kx2 + 2x + 1 = 0
The given equation is Kx2 + 2x + 1 = 0
Here, a = k, b = 2, c = 1
The discriminate (D) = b2 – 4ac = 0
= 4 – 4k = 0
= 4k = 4
K = 1
The value of k = 1 for which the quadratic equation is having real and equal roots.
(vi) Kx2 + 6x + 1 = 0
The given equation is Kx2 + 6x + 1 = 0
Here, a = k, b = 6, c = 1
The discriminate (D) = b2 – 4ac = 0
= 36 – 4k = 0
= 4k = 36
= K = 9
The value of k = 9 for which the quadratic equation is having real and equal roots.
(vii) x2 – kx + 9 = 0
The given equation is X2 – kx + 9 = 0
Here, a = 1, b = -k, c = 9
Given that the equation is having real and distinct roots. Hence, the discriminate (D) = b2 – 4ac = 0
= k2 – 4(1)(9) = 0
= k2 – 36 = 0
= k = – 6 and k = 6
The value of k lies between -6 and 6 respectively to have the real and distinct roots.
Question: 5
Find the values of k for which the given quadratic equation has real and distinct roots.
(i) Kx2 + 2x + 1 = 0
(ii) Kx2 + 6x + 1 = 0
Solution:
(i) Kx2 + 2x + 1 = 0
The given equation is Kx2 + 2x + 1 = 0
The given equation is in the form of ax2 + bx + c = 0 so, a = k, b = 2, c = 1 D = b2 – 4ac = 0
= 4 – 4(1)(k) = 0
= 4k = 4
k = 1
The value of k for the given equation is k = 1
(ii) Kx2 + 6x + 1 = 0
The given equation is Kx2 + 6x + 1 = 0
The given equation is in the form of ax2 + bx + c = 0 so, a = k, b = 6, c = 1
D = b2 – 4ac = 0
= 36 – 4(1)(k) = 0
= 4k = 36
= k = 9
The value of k for the given equation is k = 9
Question: 6
For what value of k, (4 – k)x2 + (2k + 4)x + (8k + 1) = 0, is a perfect square.
Solution:
The given equation is (4 – k)x2 + (2k + 4)x + (8k + 1) = 0
The given equation is in the form of ax2 + bx + c = 0 so, a = 4 – k, b = 2k + 4, c = 8k + 1 D = b2 – 4ac
= (2k + 4)2 – 4(4 – k)(8k + 1)
= 4k2 + 16 + 4k – 4(32 + 4 – 8k2 – k)
= 4(k2 + 4 + k – 32 – 4 + 8k2 + k)
=4(9k2 – 27k)
Since the given equation is a perfect square Therefore D = 0 = 4(9k2 – 27k) = 0
= (9k2 – 27k) = 0
= 3k(k – 3) = 0
Therefore 3k = 0
K = 0 Or, k-3 = 0 K = 3 The value of k should be 0 or 3 to be perfect square.
Question: 7
If the roots of the equation (b – c)x2 + (c – a)x + (a – b) = 0 are equal , then prove that 2b = a + c.
Solution:
The given equation is (b – c)x2 + (c – a)x + (a – b) = 0.
The given equation is the form of ax2 + bx + c = 0.
So, a = (b – c), b = (c – a), c = (a – b)
According to question the equation is having real and equal roots.
Hence discriminant (D) = b2 – 2ac = 0
= (c – a)2 – 4(b – c)(a – b) = 0
= c2 + a2 – 2ac – 4(ab – b2 – ac + cb) = 0
= c2 + a2 – 2ac – 4ab + 4b2 + 4ac – 4cb = 0
= c2 + a2 + 2ac – 4ab + 4b2 – 4cb = 0
= (a + c)2 – 4ab + 4b2 – 4cb = 0
= (c + a – 2b)2 = 0
= (c + a – 2b) = 0
= c + a = 2b
Hence it is proved that c + a = 2b.
Question: 8
If the roots of the equation (a2 + b2) x2 – 2(ac + bd)x + (c2 + d2) = 0 are equal. Prove that a ÷ b = c ÷ d.
Solution:
The given equation is (a2 + b2)x2 – 2(ac + bd)x + (c2 + d2) = 0.
The equation is in the form of ax2 + bx = c = 0
Hence, a = (a2 + b2), b = – 2(ac + bd), c = (c2 + d2).
The given equation is having real and equal roots.
Discriminant (D) = b2 – 4ac = 0
= [-2(ac + bd)]2 – 4 (a2 +b2)(c2 + d2) = 0
= (ac + bd)2 – (a2 + b2)(c2 + d2) = 0
= a2c2 + b2d2 + 2abcd – (a2c2 + a2d2 + b2c2 + b2d2) = 0
Cancelling out the equal and opposite terms.
We get, = 2abcd – a2d2 – b2c2 = 0
= abcd + abcd – a2d2 – b2c2 = 0
= ad(bc – ad) + bc(ad – bc) = 0
= ad(bc – ad) – bc(bc – ad) = 0
= (ad – bc)(bc – ad) = 0
= ad – bc = 0
= (a ÷ b) = (c ÷ d)
Hence, it is proved.
Question: 9
If the roots of the equation ax2 + 2bx + c = 0 and
are simultaneously real, then prove that b2 – ac = 0.
Solution:
The given equations are ax2 + 2bx +c = 0 and
These two equations are of the form ax2 + bx + c = 0.
Given that the roots of the two equations are real.
Hence, D = 0 that is b2 – 4ac = 0
Let us assume that ax2 + 2bx + c = 0 be equation (i) andbe (ii) From equation (i) b2 – 4ac = 0
= 4 b2 – 4ac = 0 …. (iii)
From equation (ii) b2 – 4ac = 0
Given, that the roots of equation (i) and (ii) are simultaneously real and hence equation (iii) = equation (iv).
= 4b2 – 4ac = 4ac – 4
b2 = 8ac = 8b2 = b2 – ac = 0.
Hence it is proved that b2 – ac = 0.
Question: 10
If p, q are the real roots and p = q. Then show that the roots of the equation (p-q)x2 + 5(p + q)x – 2(p – q) = 0 are real and equal.
Solution:
The given equation is (p – q)x2 + 5(p + q)x – 2(p – q) = 0
Given, p , q are real and p ? q.
Then, Discriminant (D) = b2– ac = [5(p + q)]2 – 4(p – q)(-2(p – q)) = 25(p + q)2 + (p – q)2
We know that the square of any integer is always positive that is, greater than zero.
Hence, (D) = b2 –ac = 0 As given, p, q are real and p = q.
Therefore, = 25(p + q)2 + (p – q)2 ? 0 = D = 0
Therefore, the roots of this equation are real and unequal.
Question: 11
If the roots of the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal , then prove that either a = 0 or a³ + b³ + c³ = 3abc .
Solution:
The given equation is (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0
This equation is in the form of ax2 + bx + c = 0
So, a = (c2 – ab), b = -2(a2 – bc), c = b2 – ac.
According to the question, the roots of the given question are equal.
Hence, D= 0, b2 – 4ac = 0
= [-2(a2 – bc)]2 – 4(c2 – ab)( b2 – ac) = 0
= 4(a2 – bc)2 – 4(c2 – ab)( b2 – ac) = 0
= 4a(a³ + b³ + c³ – 3abc) = 0
Either 4a =0 therefore, a = 0 Or, (a³ + b³ + c³ – 3abc) = 0
= (a³ + b³ + c³) = 3abc Hence its is proved.
Question: 12
Show that the equation 2(a2 + b2)x2 + 2(a + b)x + 1 = 0 has no real roots , when a = b.
Solution:
The given equation is 2(a2 + b2)x2 + 2(a + b)x – 1 = 0
This equation is in the form of ax2 + bx + c = 0
Here, a = 2(a2 + b2), b = 2(a + b), c = +1.
Given, a = b The discriminant (D) = b2 – 4ac = [2(a + b)]2 – 4(2(a2 + b2))(1)
= 4(a + b)2 – 8(a2 + b2)
= 4(a2 + b2 + 2ab) – 8a2 – 8b2
= +2ab – 4a2 – 4b2
According to the question a = b, as the discriminant D has negative squares so the value of D will be less than zero. Hence, D = 0, when a = b.
Question: 13
Prove that both of the roots of the equation (x – a)(x – b) +(x – c)(x – b) + (x – c)(x – a) = 0 are real but they are equal only when a = b = c.
Solution:
The given equation is (x – a)(x – b) + (x – c)(x – b) + (x – c)(x – a) = 0
By solving the equation, we get it as, 3x2 – 2x(a + b + c) + (ab + bc + ca) = 0
This equation is in the form of ax2 + bx + c = 0
Here, a = 3, b = 2(a + b + c), c = (ab + bc + ca)
The discriminate (D) = b2 – 4ac
= [-2(a + b + c)]2 – 4(3)(ab + bc + ca)
= 4(a + b +c )2 -12(ab + bc + ca)
= 4[(a + b + c)2 – 3(ab + bc + ca)]
= 4[a2 + b2 + c2 – ab – bc – ca]
= 2[2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca]
= 2[(a – b)2 + (b – c)2 + (c – a)2]
Here clearly D = 0, if D = 0 then, [(a – b)2 + (b – c)2 + (c – a)2] = 0
a –b = 0 b – c = 0 c – a = 0
Hence, a = b = c = 0 Hence, it is proved.
Question: 14
If a, b, c are real numbers such that ac = 0, then, show that at least one of the equations ax2 + bx + c = 0 and – ax2 + bx + c = 0 has real roots.
Solution:
The given equation are ax2 + bx + c = 0 … (i)
And – ax2 + bx + c = 0 … (ii)
Given, equations are in the form of ax2+ bx + c = 0 also given that a, b, c are real numbers and ac = 0.
The Discriminant (D) = b2 – 4ac For equation (i) = b2 – 4ac … (iii)
For equation (ii) = b2 – 4(-a)(c) = b2 + 4ac … (iv)
As a, b, c are real and given that ac = 0
Hence b2 – 4ac = 0 and b2+ 4ac = 0 Therefore, D = 0 Hence proved.
Question: 15
If the equation (1 + m2)x + 2mcx + (c2 – a2) = 0 has real and equal roots , prove that c2 = a2(1 + m2).
Solution:
The given equation is (1 + m2)x2 + 2mcx + (c2 – a2) = 0
The above equation is in the form of ax2 + bx + c = 0.
Here a = (1 + m2), b = 2mc, c = +(c2 – a2)
Given, that the nature of the roots of this equation is equal and hence D = 0, b2 – 4ac = 0
= (2mc)2 – 4(1 + m2)(c2 – a2) = 0
= 4m2c2 – 4(c2 + m2c2 – a2 – a2m2) = 0
= 4(m2c2 – c2 + m2c2 + a2 + a2m2) = 0
= m2c2 – c2 + m2c2 + a2 + a2m2 = 0
Now cancelling out the equal and opposite terms, = a2 + a2m2 – c2 = 0
= a2 (1 + m2) – c2 = 0
Therefore, c2 = a2 (1 + m2) Hence it is proved that as D = 0, then the roots are equal of c2 = a2 (1+ m2).
All Chapter RD Sharma Solutions For Class10 Maths
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