Simplify (radicand 5  -2 radicand 14) (2- radicand 2).The -2 is not in the radical sign.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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The request of the problem is vague, hence, if you need to perform the multiplication of binomials `(sqrt 5 - 2sqrt 14)(2 - sqrt 2)` you need to use the property of multiplication to be distributive over addition, such that:

`(sqrt 5 - 2sqrt 14)(2 - sqrt 2) = 2sqrt 5 - sqrt 5*sqrt 2 - 2*2sqrt 14 + (-2sqrt 14)(-sqrt 2)`

`(sqrt 5 - 2sqrt 14)(2 - sqrt 2) = 2sqrt 5- sqrt 10 - 4sqrt 14 + 2(sqrt(2*7*2))`

`(sqrt 5 - 2sqrt 14)(2 - sqrt 2) = 2sqrt 5- sqrt 10 - 4sqrt 14 + 4sqrt 7`

You may group the terms such that:

`(sqrt 5 - 2sqrt 14)(2 - sqrt 2) = (2sqrt 5- sqrt 10) - (4sqrt 14 - 4sqrt 7)`

`(sqrt 5 - 2sqrt 14)(2 - sqrt 2) = sqrt 5(2- sqrt 2) - 4sqrt 7(sqrt 2 - 1)`

Hence, performing the multiplication yields `(sqrt 5 - 2sqrt 14)(2 - sqrt 2) = sqrt 5(2- sqrt 2) - 4sqrt 7(sqrt 2 - 1).`

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