RD Sharma Solutions Class 10 Quadratic Equations Exercise 8.6

RD Sharma Class 10 Solutions Chapter 8 Ex 8.6 PDF Free Download

Exercise 8.6

 

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

 

Solutions: 

Important Notes: 

  • A quadratic equation is in the form ax2 +bx +c =0
  • To find the nature of roots, first find determinant “D”
  • D = b2 – 4ac
  • If D > 0, equation has real and distinct roots
  • If D < 0, equation has no real roots
  • If D = 0, equation has 1 root

(i) 2x2 -3x + 5 =0

Here, a= 2, b= -3, c= 5

D = b2 – 4ac

= (-3)2 -4(2)(5)

= 9 – 40

= -31<0

Here D<0 and so, the equation does not have any real roots.

 

(ii) 2x2 -6x + 3=0

Here, a= 2, b= -6, c= 3

D = (-6)2 -4(2)(3)

= 36 – 24

= 12<0

As D>0, the expression have real and distinct roots.

 

(iii) For what value of k (4-k)x2 + (2k+4)x +(8k+1) =0 is a perfect square.

Here, a= 4-k, b= 2k+4, c= 8k+1

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)

As the given equation is a perfect square, D= 0

So,

4(9k2-27k) = 0

⇒ 9k2-27k=0

⇒ k2 -3k =0

⇒ K (k-3) =0

Now, for this equation to be true,

k = 0 or k = 3

∴ The value of k has to be either 0 or 3 for the equation to be a perfect square.

 

(iv) Find the least positive value of k for which the equation x2+kx+4=0 has real roots.

Here, a= 1, b= k, c= 4

Since the equation has real roots,  D will be positive

D = b– 4ac ≥0

= k2 – 16 ≥ 0

⇒ k ≤ 4 ,k ≥ -4

∴ The least positive value of k  will be 4 for which the given equation will 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

Here, a= k, b= 2, c= 1

D = b2 – 4ac ≥0

= 4 -4k ≥ 0 = 4k ≤4

∴ K ≤ 1

 

(vi) Kx2 +6x+1 = 0

Here, a= k, b= 6, c= 1

D = b2 –4ac ≥0

= 36 -4k ≥ 0

= 4k ≤36

∴ K ≤ 9

 

(vii) x2 –kx+9 =0

Here, a= 1, b= -k, c= 9

It is given that the equation has real and distinct roots.

So, D = b2 – 4ac ≥0

= k2 – 4(1)(9) ≥ 0

= k2 -36 ≥ 0

⇒ k≥ -6 and k ≤6

∴ For the equation to have real and distinct roots, the value of k has to be between -6 and 6.

 

Question 2: Find the value of k.

(i) Kx2+4x+1=0

Here,

a= k, b= 4, c= 1

As the equation has real and equal roots,

D = b2-4ac = 0

= 42-4(k)(1)=0

= 16-4k=0

⇒ k = 4

 

(ii) kx− 250

Here, a= k , b= 25 , c= 4

It is given that the equation has real and equal roots so,

D = 0

Or, b2-4ac = 0

−(25)− × × 0

= 20- 16k =0

⇒ 5/4

 

(iii) 3x2-5x+2k =0

Here,

a= 3, b= -5, c= 2k

As the equation has real and equal roots, D = 0

Or, b2-4ac = 0

= (-5)2-4(3)(2k) = 0

= 25- 24k =0

K = 25/24

 

(iv) 4x2+kx+9=0

Given,

a= 4, b= k, c= 9

Here, D = 0 as the equation has real and equal roots.

⇒ b2-4ac=0

=  k2-4(4)(9)=0

= k2 -144=0

⇒ k = 12

 

(v) 2kx2-40x+25=0

Given,

a= 2k, b= -40, c= 25

Here D = 0 since the given equation has real and equal roots

∴ b2-4ac=0

= (-40)2-4(2k)(25)=0

= 1600 – 200k=0

⇒ k = 8

 

(vi) 9x2-24x+k = 0

Here,

a= 9, b= -24, c= k

As the equation has real and equal roots,

D = b2-4ac=0

Or, (-24)2-4(9)(k)=0

= 576-36k = 0

⇒ k = 16

 

(vii) 4x2-3kx+1 =0

Here,

a= 4, b= -3k, c= 1

Since the given equation has real and equal roots, D = 0

Or, b2-4ac=0

= (-3k)2-4(4)(1)=0

= 9k2-16=0

K = 4/3

 

(viii) x2-2(5+2k)x+3(7+10k) = 0

From the equation,

a= 1, b=+2(52k) , c= 3(7+10k)

Here the nature of the roots of the equation are real and equal roots.

So, D = 0

Or, 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

⇒ 2k2-5k+2=0

⇒ 2k2-4k-k+2=0

⇒ 2k(k-2)-1(k-2) =0

⇒ (k-2)(2k-1) = 0

∴ K = 2 Or k = ½

 

(ix) (3k+1)x2+2(k+1)x+k =0

Here a= 3k+1, b=+2(k+1) , c= (k)

It is given that the nature of the roots of the equation are real and equal. So, D = 0

Or, b2-4ac=0

= [2(k+1)]2 -4(3k+1)(k) =0

= (k+1)2-k(3k+1)=0

= -2k2+k+1 =0

Multiplying the equation with -1,

2k– k – 1 = 0

The value of k can be obtained by

K = \(\frac{1+\sqrt{9}}{4}\) = 1

Or , \(k=\frac{1-\sqrt{9}}{4}\)  = \(\frac{-1}{2}\)

The value of k are 1 and \(\frac{-1}{2}\)

 

(x) Kx2+kx+1 = -4x2-x

This equation has to be made in the form of a quadratic equation i.e. ax2+bx+c=0.

So, bring all the x components to LHS,

Now the equation becomes

x2(4+k)+x(k+1)+1 =0

Here a=4+k,b=+k+1 , c= 1

Now, D = 0 as the nature of the roots of the equation are real and equal

or, b2-4ac=0

= (k+1)2-4(4+k)(1) =0

= k– 2k-10 =0

It can be seen that this equation also in the form ax2+bx+c=0

So, solve the equation as a quadratic.

Here, a=1 , b= -2 , c = -15

\(k=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)

Or, k = 5 and -3

 

(xi) (k+1)x2+2(k+3)x+k+8 =0

Here,

a=k+1,b=2(k+3), c= k+8

b2-4ac =0

= [2(k+30]2-4(k+1)(k+8)=0

= 4(k+3)2-4(k+1)(k+8) =0

Take out 4 as common from the LHS and divide the same on the RHS

= (k+3)2-(k+1)(k+8) =0

⇒ k2+9+6k – ( k2+9k+18) =0

⇒ 9+6k-9k-8 =0

⇒= -3k+1 =0

⇒ 3k = 1

k = 1/3

 

 

(xii) x2-2kx+7k-12=0

 

Here 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

\(k=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)

Here a = 1 , b = -7k , c = 12

By calculating the value of k is \(\frac{7\pm \sqrt{1}}{2}\)  = 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

 

Here a=k+1,b=-2(k+1), c= 8k+1

D= b2-4ac =0

= (-2(k+1))2-4(k+1)(8k+1) =0

= 4(3k+1)2-4(k+1)(8k+1) =0

Take out 4 as common from LHS and divide the same on 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

 

(xiv) 5x2-4x+2 +k(4x2-2x+1) =0

The given equation can be written as x2(5+4k)-x(4+2k)+2-k =0

Here a = 5 + 4k, b = -(4 + 2k), c= 2 – k

As the nature of the roots of the equation are real and equal, D = 0

Or, 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

Take out 4 as common from LHS and divide the same on RHS

= 5k2-k-6 =0

The value of k can be obtained by equation

\(k=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)

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

\(k=\frac{1\pm \sqrt{-1^{2}-4(5)(-6)}}{2(5)}\)

\(k = \frac{6}{5} and -1\)

The value of k for the given equation are \(k = \frac{6}{5} and -1\) respectively.

 

(xv) (4-k)x2+(2k+4)x+(8k+1) =0

Here a=4-k, b=(2k+4), c= 8k+1

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

⇒ 36k2-108k =0

Take out 4 as common from LHS and divide the same on RHS

= 9k2-27k =0

= 9k(k-3) =0

Either 9k =0

K =0

Or, K = 3

 

(xvi) (2k+1)x2+2(k+3)x+(k+5)  = 0

Here a=2k+1, b=2(k+3), c= k+5

D= b2-4ac =0

= [2(k+3)]2-4(2k+1)(k+5) =0

Take out 4 as common from LHS and divide 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 \(k = \frac{6}{5} and -1\) respectively.

Here a= 1 , b = 5 , c= -4

Now \(k=\frac{-5\pm \sqrt{5^{2}-4(1)(-4)}}{2\times 1}\)

K = \(k=\frac{-5\pm \sqrt{41}}{2}\)

The value of k for the given equation is \(k=\frac{-5\pm \sqrt{41}}{2}\)

 

(xvii) 4x2-2(k+1)x+(k+4) =0

Here a=4, b=-2(k+1), c= k+4

D= b2-4ac =0

= [-2(k+1)]2-4(4)(k+4)=0

Take out 4 as common from LHS and divide the same on 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 \(k = \frac{6}{5} and -1\) respectively.

Here a= 1 , b = -2 , c= -15

K = \(k=\frac{2\pm \sqrt{69}}{2}\)

The value of k for the given equation is \(k=\frac{2\pm \sqrt{69}}{2}\)

 

Question 3: In the following, determine the set of values of k for which the given quadratic equation has real roots:

 

Solution:

(i) 2x2+3x+k = 0

From the equation,

a= 2 , b= 3 , c =k

Here, the given quadratic equation has equal and real roots

So, D = b2-4ac ≥ 0

= 9 – 4(2)(k) ≥ 0

= 9-8k ≥ 0

=  ≤ 9/8

The value of k does not exceed ≤ (9/8) to have a real root.

 

(ii) 2x2+kx+3 = 0

From the equation,

a= 2 , b= k , c =3

Here, D = b2-4ac ≥ 0

= k2-4(2)(3) ≥ 0

= k2-24 ≥ 0

= k2 ≥ 24

≥ 24

≥ 24

The value of k should not exceed ≥ 24 order to obtain real roots.

 

(iii)2x2-5x-k =0

Here, a= 2 , b= -5 , c =-k

AS the given quadratic equation has equal and real roots,

D = b2-4ac ≥ 0

= 25 -4(2)(-k) ≥ 0

= 25-8k ≥ 0

= k ≤ 25/8

The value of  k should not exceed ≤ 25/8

 

(iv) Kx2+6x+1 =0

Here,

a= k , b= 6 , c =1

Also,

D = b2-4ac ≥ 0 (as the quadratic equation has equal and real roots)

= 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

Here,

a= 1 , b= -k , c =9

Also, D = b2-4ac ≥ 0 (as the given equation has equal and real roots)

= k2-4(1)(-9) ≥ 0

= k2-36 ≥ 0

= k2 ≥ 36

= k ≥ 36

⇒ K ≥ 6 or k ≤ -6

 

Question 4: Determine the nature of the roots of the following quadratic equations.

Solution:

(i) 2x2 -3x + 5 =0

Here,

a= 2, b= -3, c= 5

D = b2 – 4ac

= (-3)2 -4(2)(5)

= 9 – 40

= -31 < 0

As D < 0, the equation does not having any real roots.

 

(ii) 2x2 -6x + 3=0

Here,

a= 2, b= -6, c= 3

D = b2 – 4ac

= (-6)2 -4(2)(3)

= 36 – 24

= 12<0

As D>0, the equation 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

Here, a= 4-k, b= 2k+4, c= 8k+1

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)

As the given equation is a perfect square,

D = 0

4(9k2-27k) = 0

⇒ 9k2-27k=0

⇒ K2 -3k =0

⇒ K (k-3) =0

∴ k = 0 Or, k =3

 

(iv) Find the least positive value of k for which the equation x2+kx+4=0 has real roots.

Here,

a= 1, b= k, c= 4

D = b2 – 4ac ≥0

= k2 – 16 ≥ 0

= k≤ 4  Or, k≥-4

 

(v) Find the value of k for which the given quadratic equation has real roots and distinct roots.

Kx2+ 2x+1 =0

Here,

a= k, b= 2, c= 1

D = b2 –4ac ≥0

= 4 -4k ≥ 0

= 4k ≤4

K ≤ 1

For the quadratic equation to have real and equal roots, k ≤ 1.

(vi) Kx2 +6x+1 = 0

Here,

a= k, b= 6, c= 1

D = b2 – 4ac ≥0

= 36 -4k ≥ 0

= 4k ≤36

= k ≤ 9

For the quadratic equation to have real and equal roots, k ≤ 9.

(vii) x2 –kx+9 =0

Here,

a= 1, b= -k, c= 9

As the equation is having real and distinct roots,

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 for the equation to have real and distinct roots.

 

Question 5: Find the values of k for which the given quadratic equation has real and distinct roots.

 

Solution:

(i) Kx2+2x+1 =0

Here, a= k, b= 2 , c =1

D = b2-4ac ≥ 0

= 4-4(1)(k) ≥ 0

= 4k ≤ 4

= k ≤ 1

 

(ii) Kx2+6x+1 =0

Here,

a= k, b= 6 , c =1

D = b2-4ac ≥ 0

= 36-4(1)(k) ≥ 0

= 4k ≤ 36

= k ≤ 9

 

Question 6: For what value of k , (4-k)x2+(2k+4)x+(8k+1)=0 , is a perfect square.

 

Solution:

Here,

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)

It is given that the given equation is a perfect square

∴ D =0

= 4(9k2-27k) =0

= (9k2-27k) =0

= 3k (k-3) =0

⇒ k = 0 Or, k =3

 

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:

Here,

a= (b-c), b= (c-a), c= (a-b)

It is given that the equation has real and equal roots.

So, 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

∴ It is proved.

 

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:

Here,

a= (a2 +b2) ,b= – 2(ac+bd) , c=(c2+ d2) .

It is given that the equation has real and equal roots.

So, 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

= 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, proved.

 

Question 9:  If the roots of the equation ax+ 2bx + c = 0 and bx− 2c√(ax) 0 are simultaneously real , then prove that b– ac =0.

 

Solution:

The given equations are ax2+2bx +c=0 and  bx22c√(ax) 0

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)

And bx22c√(ax) 0 be (ii)

From equation (i) b2-4ac ≥ 0

= 4 b2-4ac ≥ 0 …………………………………….. (iii)

From equation (ii) b2-4ac ≥ 0

(2√(ca))24b20……………………………….  (iv)

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.

Thus, 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 i.e. > zero.

As, D = b2 –ac ≥ 0

As p , q are real and p ≠ q.

⇒ 25(p+q)2 +(p-q)2 ˃ 0 = D ˃ 0

∴ 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:

Here,

a = (c2-ab), b = -2(a2-bc), c = b2-ac.

It is given that the roots of the given question are equal.

So, 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 ⇒ a =0

Or, (a³+b³+c³- 3abc) =0

= (a³+b³+c³) =3abc

Hence 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:

Here, a= 2(a2+b2) , b= 2(a+b) , c = +1.

Given, a ≠ b

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.

∴ 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 equation has to be solved to get the form of a quadratic equation i.e. ax2+bx+c = 0

So,

(x-a)(x-b) +(x-c)(x-b)+(x-c)(x-a) =0 ⇔ 3x2-2x(a+b+c)+(ab+bc+ca) =0

This equation is in the form of quadratic equation with a =3 , b = 2(a+b+c) , c = (ab+bc+ca)

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

Or, a = b = c = 0

Hence 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.

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

∴ 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:

Here,

a = (1+m2) , b= 2mc , c= +(c2-a2)

It is 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

Take mas common and rearrange the terms:

= a2 +a2m2 – c2 =0

= a2 (1+ m2) – c2 =0

Or, c2 = a2 (1+ m2)

Thus, it is proved that as D= 0 , then the roots are equal of c2 = a2 (1+ m2).

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