Find T if 19x^3+92x^2+23x+T=0 has three roots in geometric progression.

The only possible value of T in this equation is 19/64.

Expert Answers

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If the polynomial has three roots `a , b , c , ` then it has the form `p ( x ) = 19 ( x - a ) ( x - b ) ( x - c ) . ` It, in turn, is equal to

`19 x^3 - 19 ( a + b + c ) x^2 + 19 ( ab + bc + ac ) x - 19 abc`

and, as it is given, it is equal to

`19 x^3 + 92 x^2 + 23x + T .`

This gives three equations:

`-19 abc = T , 19 ( ab + bc + ac ) = 23 , -19 ( a + b + c ) = 92 .`

Now use the fact that `a , b , c ` must form a geometric progression. This means `b = ar ` and `c = ar^2 ` for some `r , ` and we obtain

`-19 a^3r^3 = T , 19 a^2 (r + r^2 + r^3 ) = 23 , -19 a ( 1 + r + r^2 ) = 92 .`

Divide the second equation by the third and obtain `-ar = 23 / 92 = 1 / 4 , ` i.e. `ar = -1 / 4 .`

Now the first equation gives that `T = 19 / 64 .`

We can also state that the root `b = ar ` is equal to `-1/4 .`

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