Rivalry2

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Original 2026-01-01

Modified 2026-02-28

1 When the denominator has three terms or<a>trinomials</a>, like \(a±\sqrt{b}±\sqrt{c}\), the process of rationalizing is more complex. To rationalize a trinomial denominator, treat two terms as a single<a>binomial</a>, and the third term separately. Select a rationalizing factor that simplifies at least one irrational term when multiplied. If radicals are not eliminated completely after the first rationalization process, repeat the process with the obtained result to eliminate the remaining irrational terms.

2 For instance, the given<a>expression</a>is \(\frac{1}{1 + \sqrt{2} - \sqrt{3}} \)

3 Step 1:Select two terms to form a binomial. Here, we choose \((1 + \sqrt{2}) \) and treat -√3 as the third term. So, the conjugate of the three terms is \((1 - \sqrt{2}) + \sqrt{3}\).

4 Now, we can multiply this conjugate with both the numerator and the denominator.

5 \(\frac{1}{1 + \sqrt{2} - \sqrt{3}} \times \frac{1 - \sqrt{2} + \sqrt{3}}{1 - \sqrt{2} + \sqrt{3}} \)

6 Step 2:Simplify the denominator by using the difference of squares<a>formula</a>:

7 (a - b) (a + b) = a2 - b2

8 Here,\( a = (1 + \sqrt{2})\) and \(b = \sqrt{3}\)

9 \((1 + \sqrt{2})^2 - (\sqrt{3})^2\)

10 We can expand it to:

11 \(1 + 2\sqrt{2} + 2 - 3\)

12 \(= 3 + 2\sqrt{2} - 3\)

13 \(= 2\sqrt{2}\)

14 Here the fraction becomes:

15 \(\frac{1 + \sqrt{2} + \sqrt{3}}{2 \sqrt{2}} \)

16 Step 3:Multiply by √2 in both the numerator and the denominator.

17 \(\frac{(1 + \sqrt{2} + \sqrt{3}) \cdot 2\sqrt{2} \times \sqrt{2}}{\sqrt{2}} \)

18 So, we can expand the numerator as:

19 \(\sqrt{2 }+ 2 + \sqrt{6 }\)

20 Next, we can expand the denominator as:

21 \(2\sqrt{2} × \sqrt{2} = 4 \)

22 Therefore, the final answer is \(\frac{\sqrt{2} + 2 + \sqrt{6}}{4} \).