Find the product of the pair of conjugates.

$(2\sqrt{3}-\sqrt{7})(2\sqrt{3}+\sqrt{7})$

$(2\sqrt{3}-\sqrt{7})(2\sqrt{3}+\sqrt{7})$

Marisol Rivers
2022-07-27
Answered

Find the product of the pair of conjugates.

$(2\sqrt{3}-\sqrt{7})(2\sqrt{3}+\sqrt{7})$

$(2\sqrt{3}-\sqrt{7})(2\sqrt{3}+\sqrt{7})$

You can still ask an expert for help

Clarissa Adkins

Answered 2022-07-28
Author has **16** answers

For this question, we are going to apply the same method we use inpolynomials called the FOIL Method which states:

First

Outer

Inner

Last

Therefore, let's follow this process to get the following:

$[2\sqrt{3}][2\sqrt{3}]+[2\sqrt{3}][\sqrt{7}]+[-2\sqrt{3}][\sqrt{7}]-[\sqrt{7}][\sqrt{7}]$

Now, whenever we're multiplying the same radical as seen above, wecan simplify it as just the number inside the radical beingmultiplied by the constants to get the following:

$4(3)+2\sqrt{3}\sqrt{7}-2\sqrt{3}\sqrt{7}-7$

With further simplification from the above we can get the followingas a result:

FINAL ANSWER: 12 - 7 = 5

First

Outer

Inner

Last

Therefore, let's follow this process to get the following:

$[2\sqrt{3}][2\sqrt{3}]+[2\sqrt{3}][\sqrt{7}]+[-2\sqrt{3}][\sqrt{7}]-[\sqrt{7}][\sqrt{7}]$

Now, whenever we're multiplying the same radical as seen above, wecan simplify it as just the number inside the radical beingmultiplied by the constants to get the following:

$4(3)+2\sqrt{3}\sqrt{7}-2\sqrt{3}\sqrt{7}-7$

With further simplification from the above we can get the followingas a result:

FINAL ANSWER: 12 - 7 = 5

scherezade29pc

Answered 2022-07-29
Author has **2** answers

$(2\sqrt{3}-\sqrt{7})(2\sqrt{3}+\sqrt{7})$, FOIL method, this look like$(a-b)(a-b)={a}^{2}-{b}^{2}$

$=(2\sqrt{3})(2\sqrt{3})-2\sqrt{3}\ast \sqrt{7}+2\sqrt{3}\ast \sqrt{7}-\sqrt{7}\ast \sqrt{7}$

=4(3)-7

=5

$=(2\sqrt{3})(2\sqrt{3})-2\sqrt{3}\ast \sqrt{7}+2\sqrt{3}\ast \sqrt{7}-\sqrt{7}\ast \sqrt{7}$

=4(3)-7

=5

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