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Answer no 16 and 17

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

Additional comment

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.

phosphate, calcium ion, chloride ion, neon atom

step-by-step explanation: