If is a root of a polynomial, the equation lets high powers of be rewritten in terms of lower ones. Together with Vieta’s formulas, this evaluates expressions in the roots without finding the roots.
Problem 10.16.1. Let be a root of . Find .
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Let .. Also and , from :
oops lol.
So
Problem 10.16.2. Find the sum of the 20th powers of the roots of .
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Let be a root. So
Then (on a later page):
and because , so the sum is
Problem 10.16.3. Let be the roots of (they are real). Prove that .
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? We have , so
Or:
Instead:
So since are roots, , so . Then
, so , therefore .
Example 10.16.4. has roots , and has roots . Find .
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:
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: ✗
Example 10.16.5..
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so and , and
Example 10.16.6. Let with . What is the value of ?
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, and .
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, so . Trying :. Candidates: .(unfinished)
Example 10.16.7 (Diagonal of a box). has roots . Find the length of the interior diagonal of a box with sides .
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By Vieta,
So the volume is , and is the surface area.
Expanding gives plus ,,, that is, plus . So
Let . The diagonal of the face has , and then the interior diagonal has . So
Problem 10.16.8. Consider all lines that meet the graph of in four distinct points ,. Show that
is independent of the line and find its value.
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Let’s look at the line that intersects at . For that line, the are the roots of , so
And at height :
Exercises
Exercise 10.16.9. Let be a root of . Evaluate .
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, so
Exercise 10.16.10. Let be a root of .
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, so .
(a) , so
So .
(b) , so . So , so .
(c)
Exercise 10.16.11. Find the sum of the 20th powers of the roots of .