Ex-8.6 Quadratic Equations (Part - 2), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

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

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 ≤ 9/8

The value of k does not exceed k ≤ 9/8 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

= k²  ≥ 24

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≤258

The value of  k should not exceed k≤258

(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 ≥ 

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.

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

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 bx²− + b = 0 are simultaneously real , then prove that b²-ac = 0.

Solution:

The given equations are ax² + 2bx + c = 0 and  bx²−+ b = 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 bx²− + b = 0 be (ii)

From equation (i) b²-4ac ≥ 0

= 4 b²-4ac ≥ 0 …………………………………….. (iii)

From equation (ii) b²-4ac ≥ 0

=  − 4b²  ≥ 0……………………………….  (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.

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 discriminat (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).