Question

heya guys ans this........... plssssssssssssssss

Answer 1

$\huge\boxed{Answer}$

Let the zeroes of the cubic polynomial

3x³ - 22x² + 48x - 32 = 0

are alpha , alpha + beta , alpha + 2beta

whose are in H. P.

then , 1/alpha , 1/(alpha + beta) &

1/(alpha + 2beta) are in A.P.

Now , 1/a + 1/(a + b) + 1/(a + 2b) = 22/3

(a + b)(a + 2b) + a(a + 2b) + a(a + b)/a(a + b)(a + 2b) = 22/3 -------(1)

1/ a(a + b)(a + 2b) = 32/3 -------(2)

From eq. (1) & (2)

(a + b)(a + 2b) + a(a + 2b) + a(a + b)* 32 = 22

(a + b)(a + 2b) + a(a + 2b) + a(a + b) = 11/16 --(3)

Now ,

1/a(a + b) + 1/(a + b)(a + 2b) + 1/a(a + 2b) = 16

(a + 2b) + a + (a + b) / (3/32) = 16

(3a + 3b) * 32/3 = 16

3(a + b) * 32/3 = 16

(a + b) = 1/2 -------(4)

From eq. (3) -

1/2(1/2 + b) + (1/2 - b)(1/2 + b) + 1/2(1/2 - b) = 11/16

1/2 * (1) + ( 1/4 - b² )= 11/16

3/4 - b² = 11/16

12 - 16b² = 11

16b² = 1

b² = 1/16

b = 1/4

then , a = 1/4

then roots are :- 1/(1/4) , 1/(1/2) & 1/(3/4)

4 , 2 & 4/3

Answer 2

Let the zeroes of the cubic polynomial

3x³ - 22x² + 48x - 32 = 0

are alpha , alpha + beta , alpha + 2beta

whose are in H. P.

then , 1/alpha , 1/(alpha + beta) &

1/(alpha + 2beta) are in A.P.

Now , 1/a + 1/(a + b) + 1/(a + 2b) = 22/3

(a + b)(a + 2b) + a(a + 2b) + a(a + b)/a(a + b)(a + 2b) = 22/3 -------(1)

1/ a(a + b)(a + 2b) = 32/3 -------(2)

From eq. (1) & (2)

(a + b)(a + 2b) + a(a + 2b) + a(a + b)* 32 = 22

(a + b)(a + 2b) + a(a + 2b) + a(a + b) = 11/16 --(3)

Now ,

1/a(a + b) + 1/(a + b)(a + 2b) + 1/a(a + 2b) = 16

(a + 2b) + a + (a + b) / (3/32) = 16

(3a + 3b) * 32/3 = 16

3(a + b) * 32/3 = 16

(a + b) = 1/2 -------(4)

From eq. (3) -

1/2(1/2 + b) + (1/2 - b)(1/2 + b) + 1/2(1/2 - b) = 11/16

1/2 * (1) + ( 1/4 - b² )= 11/16

3/4 - b² = 11/16

12 - 16b² = 11

16b² = 1

b² = 1/16

b = 1/4

then , a = 1/4

then roots are :- 1/(1/4) , 1/(1/2) & 1/(3/4)

4 , 2 & 4/3