Question 710429: The roots of 2x^3 +21X ^2 +mx + q =0 are in geometric progression with a ratio of 2. Find the values of m and q and the roots Answer by jsmallt9(3758) (Show Source):
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Let's say that a root is "r". Because of the geometric progression we are given, the other two roots could be expressed as 2r and 4r. We can write an equation for a polynomial with these roots:
We now multiply this out:
Adding like terms:
This is a general equation for a polynomial whose roots are r, 2r and 4r. Our equation:
must fit this pattern. Now we find the a, m and q that make our equation fit the pattern of the general equation. We have a term of 2x^3 and the general form has only one term, . So a must be 2. Replacing the "a" in the general form with 2 we get:
which simplifies as follows:
Now the modified general form has one term, . This must match the term of our equation, . Using this we can find that r must be -3/2 in order for to be equal to . Replacing the r in our modified general form with -3/2:
Simplifying...
We can now see that "m" must be 63 and "q" must be 54.
The only thing left is the roots. We have already found that one root, r, is -3/2. The other two roots were 2r and 4r. Using -3/2 for we will find that the other roots are -3 and -6.