Question

12.
If the roots of the equation 8x3 - 14x2 + 7x-1=0 are in GP then the roots are
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Answer 1

Answer:

Step-by-step explanation:

8x³ - 14x² + 7x - 1 = 0  

8x³ - (6x² + 8x²) + (x + 6x) - 1 = 0  

8x³ - 6x² - 8x² + x + 6x - 1 = 0  

8x³ - 6x² + x - 8x² + 6x - 1 = 0  

[8x³ - 6x² + x] - [8x² - 6x + 1] = 0  

x.[8x² - 6x + 1] - [8x² - 6x + 1] = 0  

(x - 1).[8x² - 6x + 1] = 0  

(x - 1).[8x² - (4x + 2x) + 1] = 0  

(x - 1).[8x² - 4x - 2x + 1] = 0  

(x - 1).[(8x² - 4x) - (2x - 1)] = 0  

(x - 1).[4x.(2x - 1) - (2x - 1)] = 0  

(x - 1).(2x - 1).(4x - 1) = 0  

First case: (x - 1) = 0 → x - 1 = 0 → x = 1  

Second case: (2x - 1) = 0 → 2x - 1 = 0 → 2x = 1 → x = 1/2  

Third case: (4x - 1) = 0 → 4x - 1 = 0 → 4x = 1 → x = 1/4