Chapter 5: Problem 34

By multiplying both numerator and denominator of $\frac{1}{a+b \sqrt{c}}$ by $a-b \sqrt{c}$ show that $$ \frac{1}{a+b \sqrt{c}}=\frac{a-b \sqrt{c}}{a^{2}-b^{2} c} $$ Use this approach to show that $$ \frac{1}{2+\sqrt{3}}=2-\sqrt{3} $$

Short Answer

Expert verified

Answer: The simplified form of the fraction is $\frac{a-b\sqrt{c}}{a^2-b^2c}$. When applied to the example, we get $\frac{1}{2+\sqrt{3}} = 2-\sqrt{3}$.

Step by step solution

Multiply the numerator and the denominator by $(a-b\sqrt{c})$

We multiply both the numerator and the denominator of $\frac{1}{a+b\sqrt{c}}$ by $a-b\sqrt{c}$: $$ \frac{1}{a+b\sqrt{c}}\times\frac{a-b\sqrt{c}}{a-b\sqrt{c}} $$

Simplify the result

We perform the multiplication and simplify the fraction obtained in Step 1: $$ \frac{1(a-b\sqrt{c})}{(a+b\sqrt{c})(a-b\sqrt{c})} $$ $$ \frac{a-b\sqrt{c}}{a^2-(b\sqrt{c})^2} $$ By squaring $b\sqrt{c}$, we obtain: $$ \frac{a-b\sqrt{c}}{a^2-b^2c} $$

Apply the approach to the specific case

We apply the approach to the specific example of $\frac{1}{2+\sqrt{3}}$. In this case, we have $a = 2$, $b = 1$, and $c = 3$. We can then apply the result obtained in Step 2: $$ \frac{2-1\sqrt{3}}{(2)^2-(1)^2(3)} $$ $$ \frac{2-\sqrt{3}}{4-3} $$ $$ 2-\sqrt{3} $$ Thus, we have shown that $$ \frac{1}{2+\sqrt{3}} = 2-\sqrt{3}. $$

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By multiplying both numerator and denominator of \(\frac{1}{a+b \sqrt{c}}\) by \(a-b \sqrt{c}\) show that $$ \frac{1}{a+b \sqrt{c}}=\frac{a-b \sqrt{c}}{a^{2}-b^{2} c} $$ Use this approach to show that $$ \frac{1}{2+\sqrt{3}}=2-\sqrt{3} $$

Short Answer

Step by step solution

Multiply the numerator and the denominator by \((a-b\sqrt{c})\)

Simplify the result

Apply the approach to the specific case

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