9514 1404 393
Eexplanation:
16. Suppose the roots are α and kα. Then we can write the equation as ...
a(x -α)(x -kα) = 0
ax² -ax(α +kα) +akα² = 0
Comparing to the original equation, we can equate coefficients to get ...
Solving the first for α gives ...
α = -b/(a(1+k)
Substituting into the second, we have ...
c = ak(-b/(a(1+k)))²
Multiplying by a(1+k)², we get ...
(1+k)²ac = kb²
Using k=2 gives ...
9ac = 2b² . . . . . as required
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17. Using the previous result with k=1 (equal roots), we have ...
(1+k)²ac = kb²
4ac = b² . . . . . for k=1
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<em>Additional comment</em>
We observed that the problems were similar, but had different factors relating the roots. So, we elected to solve the general case, then fill in the specific values for the two problems.
