Question

find the roots of the equation 8x³+18x²-27x-27,if the roots of the equation are geometric progression
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Answer 1

Answer:

Roots are -3 / 4 , 3 /2 and (-3) .

Step-by-step explanation:

Given--->

8x³ + 18x² - 27x - 27 = 0

Roots of equation are in geometric progression

Solution--->

8x³ +18 x² -27x - 27 = 0

Let roots of equation are

α / r , α and αr

Now

product of roots

= - constant term / coefficient of x³

α/r × α × αr = - (-27 ) / 8

r is Cancel out from numerator and denominator

α³ = ( 3 / 2 )³

Taking cube root of both sides

α = 3 / 2

Now

Sum of roots

=- coefficient of x² / coefficient of x³

=> α/r + α + αr = - 18 / 8

=> 3/2 × ( 1/r + 1 + r ) = -18/8

=> (1 + r + r² ) / r = - 3 / 2

=> 2r² + 2r + 2 = - 3 r

=> 2r² + 5r + 2 = 0

Now we factroize it by splitting the middle term

=> 2r² + ( 4 +1 ) r + 2 = 0

=> 2r² + 4r + r +2 = 0

=> 2r ( r + 2 ) + 1 ( r + 2 ) = 0

=> ( r + 2 ) ( 2r + 1 ) = 0

If r + 2= 0

r = -2

So

3/2

α / r = ------- = -3 / 4

-2

α = 3 / 2

αr = 3/2 (-2) = - 3

If 2r + 1 = 0

2r = -1

r = -1 / 2

So

3/2

α / r = ----------- = -3

- 1 /2

α = 3 /2

αr = ( 3/2) ( -1/2) = -3/4

So roots are -3 / 4 , 3/2 and -3