Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
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Exercise 49 Page 350

5-sqrt(3), sqrt(2)

Practice makes perfect

We are told that the given roots are roots of a polynomial function P(x) that has rational coefficients. 5+sqrt(3) and - sqrt(2)To find two additional roots of P(x)=0, let's recall the first part of the Conjugate Root Theorem.

If P(x) is a polynomial with rational coefficients, then any irrational roots of P(x)=0 occur in conjugate pairs.

The above statement tell us that if a + sqrt(b) is an irrational root, then a - sqrt(b) is also a root. Let's use this to find an additional irrational root.

Hypotheses Conclusion
P(x) has rational coefficients 5-sqrt(3) is also a root of P(x)
5+sqrt(3) is an irrational root of P(x)=0
P(x) has rational coefficients sqrt(2) is also a root of P(x)
- sqrt(2) is an irrational root of P(x)=0

The two additional roots that we can know for certain using the Conjugate Root Theorem are 5-sqrt(3) and sqrt(2).