Ex-8.6 Quadratic Equations (Part - 1), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Question 1: Determine the nature of the roots of the following quadratic equations.

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) = b^{2 –} 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 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) = b^{2 –} 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) = b^{2 –} 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) = b^{2 –} 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) = b^{2 –} 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

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) kx²−2√5x + 4 = 0

The given equation kx²−2√5x + 4 = 0 is 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

=  −2√5² −4 × k × 4 = 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 =  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) = 0K = 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 = 0is 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 =  k =

Or , 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   = 4 , 3

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

Here a = 5 , b = -1 , c = -6

k = 6/5 and−1

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

Now k =

K =  k =

The value of k for the given equation is k =

(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

K =  k =

The value of k for the given equation is k =