Verified Answer :

olemakpadu

We try to represent each number inside the square root as a product of a square and another number.

a) 7√32 - 5√2 + √8

√32  = √(16 2) = √16 √2  = 4 √2 =  4√2

√8 = √(4 2) = √4 √2  = 2 √2 = 2√2

7√32 - 5√2 + √8 = 7(4√2) - 5√2 + 2√2 =
      
                         = 28√2  - 5√2 + 2√2          Factorize out √2

                           = (28 - 5 + 2)√2  

                          =  25√2      

b)      2√150 - 4√54 + 6√24       

√150 = √(25 6) = √25 √6 = 5√6 = 5√6

√54 = √(9 6) = √9 √6 = 3√6 = 3√6

√24 = √(4 6) = √4 √6 = 2√6 = 2√6


2√150 - 4√54 + 6√24 = 2(5√6) - 4(3√6) + 6(2√6)

                                = 25√6 - 43√6 + 62√6

                                = 10√6  - 12√6 + 12√6        Factorize √6

                               =  (10 - 12 + 12)√6

                               =  10√6

Exvited

\(7 \sqrt{32} - 5 \sqrt{2} + \sqrt{8} \\ \\ 7 \times 4 \sqrt{2} - 5 \sqrt{2} + 2 \sqrt{2} \\ \\ 28 \sqrt{2} - 5 \sqrt{2} + 2 \sqrt{2} \\ \\ 25 \sqrt{2} \\ \\\)

The final result is: 25√2 or 35.3553.

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\(2 \sqrt{150} - 4 \sqrt{54} + 6 \sqrt{24} \\ \\ 2 \times 5 \sqrt{6} - 4 \sqrt{54} + 6 \sqrt{24} \\ \\ 2 \times 5 \sqrt{6} - 4 \times 3 \sqrt{6} + 6 \sqrt{24} \\ \\ 2 \times 5 \sqrt{6} - 4 \times 3 \sqrt{6} + 6 \times 2 \sqrt{6} \\ \\ 10 \sqrt{6} - 4 \times 3 \sqrt{6} + 6 \times 2 \sqrt{6} \\ \\ 10 \sqrt{6} - 12 \sqrt{6} + 6 \times 2 \sqrt{6} \\ \\ 10 \sqrt{6} - 12 \sqrt{6} + 12 \sqrt{6} \\ \\ 10 \sqrt{6} \\ \\\)

The final result is: 10√6 or 24.4949.

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