Math Problem Statement

Rationalize the denominator of $\displaystyle \frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{7}}$, and write your answer in the form[ \frac{A\sqrt{2} + B\sqrt{3} + C\sqrt{7} + D\sqrt{E}}{F}, ]where everything is in simplest radical form and the fraction is in lowest terms, and $F$ is positive. What is $A + B + C + D + E + F$?

Solution

We are tasked with rationalizing the denominator of the expression

12+3+7,\frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{7}},

and expressing the result in the form

A2+B3+C7+DEF.\frac{A\sqrt{2} + B\sqrt{3} + C\sqrt{7} + D\sqrt{E}}{F}.

Step 1: Multiply by the Conjugate

To rationalize the denominator, we multiply both the numerator and denominator by the conjugate of 2+3+7\sqrt{2} + \sqrt{3} + \sqrt{7}. The conjugate in this case is 2+37\sqrt{2} + \sqrt{3} - \sqrt{7}. This will help eliminate the square roots in the denominator when simplified.

Thus, we multiply the expression by

2+372+37.\frac{\sqrt{2} + \sqrt{3} - \sqrt{7}}{\sqrt{2} + \sqrt{3} - \sqrt{7}}.

This gives

12+3+72+372+37=2+37(2+3+7)(2+37).\frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{7}} \cdot \frac{\sqrt{2} + \sqrt{3} - \sqrt{7}}{\sqrt{2} + \sqrt{3} - \sqrt{7}} = \frac{\sqrt{2} + \sqrt{3} - \sqrt{7}}{(\sqrt{2} + \sqrt{3} + \sqrt{7})(\sqrt{2} + \sqrt{3} - \sqrt{7})}.

Step 2: Simplify the Denominator

Now, we simplify the denominator. This is a difference of squares:

(2+3+7)(2+37)=(2+3)2(7)2.(\sqrt{2} + \sqrt{3} + \sqrt{7})(\sqrt{2} + \sqrt{3} - \sqrt{7}) = (\sqrt{2} + \sqrt{3})^2 - (\sqrt{7})^2.

First, calculate (2+3)2(\sqrt{2} + \sqrt{3})^2:

(2+3)2=(2)2+223+(3)2=2+26+3=5+26.(\sqrt{2} + \sqrt{3})^2 = (\sqrt{2})^2 + 2\cdot\sqrt{2}\cdot\sqrt{3} + (\sqrt{3})^2 = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}.

Now, subtract (7)2(\sqrt{7})^2:

5+267=2+26.5 + 2\sqrt{6} - 7 = -2 + 2\sqrt{6}.

Thus, the denominator simplifies to 2+26-2 + 2\sqrt{6}.

Step 3: Simplify the Expression

We now have:

2+372+26.\frac{\sqrt{2} + \sqrt{3} - \sqrt{7}}{-2 + 2\sqrt{6}}.

We can factor out a 2 from the denominator:

2+372(1+6).\frac{\sqrt{2} + \sqrt{3} - \sqrt{7}}{2(-1 + \sqrt{6})}.

Now, multiply both the numerator and denominator by 16-1 - \sqrt{6} (the conjugate of 1+6-1 + \sqrt{6}) to further simplify:

(2+37)(16)2((1+6)(16)).\frac{(\sqrt{2} + \sqrt{3} - \sqrt{7})(-1 - \sqrt{6})}{2((-1 + \sqrt{6})(-1 - \sqrt{6}))}.

Step 4: Simplify the Denominator Again

The denominator becomes:

(1+6)(16)=(1)2(6)2=16=5.(-1 + \sqrt{6})(-1 - \sqrt{6}) = (-1)^2 - (\sqrt{6})^2 = 1 - 6 = -5.

Thus, the denominator is 2(5)=102(-5) = -10.

Step 5: Expand the Numerator

We now expand the numerator (2+37)(16)(\sqrt{2} + \sqrt{3} - \sqrt{7})(-1 - \sqrt{6}):

=2(16)+3(16)7(16)= \sqrt{2}(-1 - \sqrt{6}) + \sqrt{3}(-1 - \sqrt{6}) - \sqrt{7}(-1 - \sqrt{6}) =226336+7+76.= -\sqrt{2} - \sqrt{2}\sqrt{6} - \sqrt{3} - \sqrt{3}\sqrt{6} + \sqrt{7} + \sqrt{7}\sqrt{6}.

Simplifying, this becomes:

262363+7+67.-\sqrt{2} - \sqrt{6}\sqrt{2} - \sqrt{3} - \sqrt{6}\sqrt{3} + \sqrt{7} + \sqrt{6}\sqrt{7}.

Step 6: Final Expression

The overall expression becomes:

[ \frac{-\sqrt{2} - \sqrt{6}\sqrt{2} - \sqrt{3} - \sqrt{6}\sqrt{3} + \sqrt{7} + \sqrt

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Math Problem Analysis

Mathematical Concepts

Algebra
Radical Expressions
Conjugates
Simplification

Formulas

Difference of squares: (a + b)(a - b) = a^2 - b^2
Multiplication of binomials: (x + y)^2

Theorems

Difference of squares theorem

Suitable Grade Level

Grades 10-12