Math, asked by ryanidiculathomas, 11 months ago

Answer these questions

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Answered by anindyaadhikari13
3

Here is your answer.

Question number b:-

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

 = 6 \sqrt{2}  + 4

Question number c:-

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

 = 2 \sqrt{7}  + 21

Question number d:-

 \sqrt{2} ( \sqrt{32}  -  \sqrt{8} )

 =  \sqrt{64}  -  \sqrt{16}

 = 8 - 4

 = 4

Answered by Anonymous
1

1) √2(6+2√2)

 \red{ \sf{ \sqrt{2} (6 + 2 \sqrt{2} )}}

 \sf multiply \: each \:  \underline {\orange { term}} \: in \: the \: parantheses \: by \sqrt{2}

 \red{ \sf{  6 \sqrt{2} +  \sqrt{2}   \times2 \sqrt{2}  }}

 { \sf{  6 \sqrt{2} +   \red{ \sqrt{2}   \times2 \sqrt{2}  }}}

 \sf calculate \: the \:  {\orange  {\underline{product.}}}

6 \sqrt{2} +   \red{  4}

 \sf  \red{solution}

6 \sqrt{2 }  + 4

 \sf  \red{alternate \: form}

 ≈12.48528

2) √7(2+3√7)

 \red{ \sf{ \sqrt{7} (2 + 3 \sqrt{7} )}}

 \sf multiply \: each \:  \underline {\orange { term}} \: in \: the \: parantheses \: by \sqrt{7}

 \red{ \sf{  2\sqrt{7} +  \sqrt{7}   \times3 \sqrt{7}  }}

 { \sf{  2\sqrt{7} +   \red{ \sqrt{7}   \times3 \sqrt{7}  }}}

 \sf calculate \: the \:  {\orange  {\underline{product.}}}

2 \sqrt{7} +   \red{  21}

 \sf  \red{solution}

2 \sqrt{7} +     21

 \sf  \red{alternate \: form}

 ≈26.2915

3) √2(√32-√8)

 \sqrt{2} ( \red{  \sqrt{32}  -  \sqrt{8}} )

 \sf simplfy \: the \:  \orange {\underline{  radical \: expression}}

 \sqrt{2} ( \red{ 4 \sqrt{2}   - 2 \sqrt{2} } )

 \sf collect \:   \orange {\underline{  like \: terms}}

 \sqrt{2} \times  \red{  2 \sqrt{2}  }

  \red{\sqrt{2} }\times   2  \red{ \sqrt{2}  }

 \sf when \: a \: \orange {\underline{  square \: root}} \: of \: an \: expression  \\  \sf is \: multiplied \: by \: itself, \: the \: result   \\  \sf is \: that \: expression.

 \red{2 }\times 2

 \red{2 \times 2}

 \sf multiply \: terms

 \red{4}

 \sf  \red{solution}

4

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