simplify 5 square root of 7 + 12 square root of 6 - 10 square root of 7 - 5 square root of 6.

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simplify 5 square root of 7 + 12 square root of 6 - 10 square root of 7 - 5 square root of 6.

Mathematics
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i need help please
hint: we can factor out square root of 7 between the first and the third term, furthermore, we can factor out square root of 6 between the second and fourth term, so we can write the subsequent step: \[\Large \begin{gathered} 5\sqrt 7 + 12\sqrt 6 - 10\sqrt 7 - 5\sqrt 6 = \hfill \\ \hfill \\ = \sqrt 7 \left( {5 - 10} \right) + \sqrt 6 \left( {12 - 5} \right) = ... \hfill \\ \end{gathered} \]
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honestly i dont know mate im really confused sorry... @Michele_Laino
following the rules of algebra of radicals, you can only sum similar radicals, namely radicals which have the same square roots as in your case
now, \[5\sqrt 7 ,\; - 10\sqrt 7 \] are similar since they both have square root of 7, right?
i was trying to work it out and the answer i got was 5 square root of 7 - 7 quare root of 6
I got this since 5-10= -5, we have: \[\sqrt 7 \left( {5 - 10} \right) = - 5\sqrt 7 \]
furthermore 12-5=7, so we can write: \[\sqrt 6 \left( {12 - 5} \right)\]
oops.. \[\sqrt 6 \left( {12 - 5} \right) = 7\sqrt 6 \]
and that will be 7 square root of 6 - 5 square root of 7 @Michele_Laino
that's right!
thank you so much can you help me with one more? @Michele_Laino
ok!
simplify square root of 5 (10-4 square root of 2) @Michele_Laino
we have to apply the distributive property of multiplication over addition, so we can write this: \[\Large \sqrt 5 \left( {10 - 4\sqrt 2 } \right) = 10\sqrt 5 - 4\sqrt 5 \sqrt 2 \]
am I right?
@Michele_Laino is that the answer to the question?
no, we have to write another step
ohhhhhh okay well as i simplify it i got 5 square root of 2 - 4 square root of 10 @Michele_Laino
more precisely we can write this: \[\sqrt 5 \sqrt 2 = \sqrt {5 \times 2} = \sqrt {10} \]
yea thats what i got @Michele_Laino
so we get: \[\sqrt 5 \left( {10 - 4\sqrt 2 } \right) = 10\sqrt 5 - 4\sqrt {10} \]
furthermore we can note that: \[10 = \sqrt {10} \sqrt {10} \]
so we can write: \[\sqrt 5 \left( {10 - 4\sqrt 2 } \right) = 10\sqrt 5 - 4\sqrt {10} = \sqrt {10} \sqrt {10} \sqrt 5 - 4\sqrt {10} \]
finally, I factor out sqrt(10) and I get: \[\begin{gathered} \sqrt 5 \left( {10 - 4\sqrt 2 } \right) = 10\sqrt 5 - 4\sqrt {10} = \sqrt {10} \sqrt {10} \sqrt 5 - 4\sqrt {10} \hfill \\ = \sqrt {10} \left( {\sqrt {10} \sqrt 5 - 4} \right) \hfill \\ \end{gathered} \]
its not too difficult just follow what the question is trying to tell you.
recalling taht: \[\sqrt {10} \sqrt 5 = \sqrt {10 \times 5} = \sqrt {50} \] we have: \[\begin{gathered} \sqrt 5 \left( {10 - 4\sqrt 2 } \right) = 10\sqrt 5 - 4\sqrt {10} = \sqrt {10} \sqrt {10} \sqrt 5 - 4\sqrt {10} \hfill \\ \hfill \\ = \sqrt {10} \left( {\sqrt {10} \sqrt 5 - 4} \right) = \sqrt {10} \left( {\sqrt {50} - 4} \right) \hfill \\ \end{gathered} \]
there is another way to simplify your radical
I start from the initial expression: \[\Large \sqrt 5 \left( {10 - 4\sqrt 2 } \right)\]
and I factor out a 2 inside the parentheses: \[\Large 10 - 4\sqrt 2 = 2\left( {5 - 2\sqrt 2 } \right)\]
so I can write: \[\Large \sqrt 5 \left( {10 - 4\sqrt 2 } \right) = \sqrt 5 \cdot 2\left( {5 - 2\sqrt 2 } \right)\]
and we have finished
now you can choose the method which do you prefer

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