Can someone help me solv this.What is (3+sqrt(5)) (7-2sqrt(5))^2 a+b sqrt(5)
I'm not sure what this equation is suppose to look like. Do you mind using the 'equation' option in comments to write it out? :)
\[(3+\sqrt{5}) (7-2\sqrt{5})^{2}\]
Thanks! Okay, just give me a minute please..
Is this suppose to be simplified?
Yes a+b \[\sqrt{5}\]
Okay well by simplifying, this is my answer.. \[67-35\sqrt{5}+4\times5^{\frac{ 3 }{ 2 }}\]
But I don't think that's what they are looking for. Do you have multiple choices?
all I got is the answer can be written as \[a+b \sqrt{5}\]
I've got it\[-15+67\sqrt{5}\]
Simplify \[(3+\sqrt{5}) (7-2\sqrt{5})^{2}\] The answer can be written as \[a+b \sqrt{5}\] I need to find a and b I tried to use \[(a+b)^{2} = a^{2}+ 2ab +b^{2}\] Messed it up somewhere and not sure how to solv it :/
I can show you the steps on how I got that answer but it's pretty long
Yes please, that would be super! :) Forever thankful!
I'm going to put it in each step at a time, okay?\[(\sqrt{5}+3)((-2\times \sqrt{5})^{2}+2(-2\times \sqrt{5})7+7^{2})\]
\[(\sqrt{5}+3)((2\times \sqrt{5})^{2}-2\times2\times \sqrt{5}\times7+49)\]
\[(\sqrt{5}+3)(2^{2}(\sqrt{5})^{2}-2\times2\times7\times \sqrt{5}+49)\]
\[(\sqrt{5}+3)(4\times \sqrt[2]{5}-28\times \sqrt{5}+49)\]
\[(\sqrt{5}+3)(4\times5-28\times \sqrt{5}+49)\]
\[(\sqrt{5}+3)(-28\times \sqrt{5}+69)\]
\[-5\times28\times \sqrt{5}+\sqrt{5}\times69\times3\times28\times \sqrt{5}+3\times69\]
\[-28\times \sqrt[2]{5\times5}+69\times \sqrt{5}-84\times \sqrt{5}+207\]
\[-28\times \sqrt[2]{5^{1+1}}+69\times \sqrt{5}-84\times \sqrt{5}+207\]
Almost done, I promise
\[-28\times \sqrt[\frac{ 2 }{ 2 }]{5^{\frac{ 2 }{ 2 }}}+69\times \sqrt{5}-84\times \sqrt{5}+207\]
\[-28\times5+69\times \sqrt{5}-84\times \sqrt{5}+207\]
\[-140+69\times \sqrt{5}-84\times \sqrt{5}+207\]
\[-15\times \sqrt{5}+67\] Put ^^ in the correct form\[(a+b \sqrt{x})\] and.. \[-15+67\sqrt{5}\] Ta-da!! :)
Sorry that was super long :(
Thank you so much! And yes that was long, thanks for doing it, did you follow a specifc formula? What ever I did doesnt quite look like that. Thanks again. :)
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