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`        IITians need your help pls....?.................................................`
one year ago

Samyak Jain
331 Points
```							54x3 – 39x2 – 24x + 16 = 0 . Roots of this equation are in GP.So let the roots be a/r , a , ar , where a/r is the first term and r is the common ratio of the GP.Sum of roots = a/r + a + ar = – (–39) /54 = 39/54        ...(1)Product of roots = (a/r)(a)(ar) = – (16) / 54  $\dpi{100} \Rightarrow$ a3 = – 16/54 = – 8/27i.e. a3 = (– 2/3)3  or  a = – 2/3From (1), a(1/r + 1 + r) = 39/54  $\dpi{100} \Rightarrow$  (– 2/3)(r2 + r + 1)/r  =  13/18.(r2 + r + 1)/r  =  – 13/12  $\dpi{100} \Rightarrow$  (r2 + r + 1) = – (13/12)r  or  r2 + (25/12)r + 1 = 012r2 + 25r + 12 = 0  i.e.  12r2 + 16r + 9r + 12 = 0  or  (3r + 4)(4r + 3) = 0r = – 4/3  or  r = – 3/4When r = – 4/3, a/r = (– 2/3)/(– 4/3) = 1/2 , a = – 2/3 , ar = (– 2/3)(– 4/3) = 8/9.When r = – 3/4, a/r = (– 2/3)/(– 3/4) = 8/9 , a = – 2/3 , ar = (– 2/3)(– 3/4) = 1/2.$\dpi{100} \therefore$ roots of the equation are 1/2 , – 2/3 , 8/9.
```
8 months ago
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