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What is the product of the radical expression?


(7 _sqrt7)(_6 +sqrt7)


a . _49 _ 42 sqrt7

b . _49 + 13 sqrt7

c . _35 _ 42 sqrt7

d . _6 + sqrt7

Verified Answer :

Nirina7

(7 _sqrt7)(_6 +sqrt7)= _49 + 13 sqrt7

Answer:

option (b) is correct

The product of \((7-\sqrt{7} )\) and \((-6+\sqrt{7} )\) is \(-49+13\sqrt{7}\)

Step-by-step explanation:

Given radicals \((7-\sqrt{7} )\) and \((-6+\sqrt{7} )\)

WE have to find the product of two given radicals

Consider \(\left(7-\sqrt{7}\:\right)\left(-6+\sqrt{7}\:\right)\)

Using property of algebra,

\(\left(a+b\right)\left(c+d\right)=ac+ad+bc+bd\) , We have ,

\(=7\left(-6\right)+7\sqrt{7}+\left(-\sqrt{7}\right)\left(-6\right)+\left(-\sqrt{7}\right)\sqrt{7}\)

Simplifying further , we get,

\(=-7\cdot \:6+7\sqrt{7}+6\sqrt{7}-\sqrt{7}\sqrt{7}\)

Adding similar terms , we get,

\(=-49+13\sqrt{7}\)

Thus, the product of \((7-\sqrt{7} )\) and \((-6+\sqrt{7} )\) is \(-49+13\sqrt{7}\)

Thus, option (b) is correct