Question

For Individual or Group Work (Exercises 147 – 150)In calculus, it is sometimes desirable to rationalize a numerator. To do this, we multiply the numerator and the denominator by the conjugate of the numerator. For example, (6 - √2)/4 = (6 - √2)/4 × (6 + √2)/(6 + √2) = (36 - 2)/(4(6 + √2)) = 34/(4(6 + √2)) = 17/(2(6 + √2)) = 17/(6 + √2).

2√10 + √7 30

Accepted Answer

Identify the expression to rationalize: $$\frac{6 - \sqrt{3}}{8}. $$The numerator is $$6 - \sqrt{3}$$, which contains a square root that we want to eliminate by rationalizing. Find the conjugate of the numerator. The conjugate of $$6 - \sqrt{3} is$$ $$6 + \sqrt{3}. $$This is because conjugates have the same terms but opposite signs between them. Multiply both the numerator and the denominator by the conjugate of the numerator to keep the expression equivalent. This gives: $$\frac{6 - \sqrt{3}}{8} 	imes \frac{6 + \sqrt{3}}{6 + \sqrt{3}}. $$Use the difference of squares formula to simplify the numerator: $$(a - b)(a + b) = a^2$ - b^2$. Here, a = 6$ and b$$ = \sqrt{3}$$, so the numerator becomes 6^2$ - (\sqrt{3})^2$. Simplify the numerator and write the new expression as $$\frac{36 - 3}{8(6 + \sqrt{3})}$$. This is the rationalized form with the numerator free of square roots.

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Conjugates in Algebra

Difference of Squares Formula