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anonymous
 one year ago
WILL MEDAL AND FAN! PLEASE, THIS IS URGENT!
Solve √(k7) = √(k) 1 for k.
anonymous
 one year ago
WILL MEDAL AND FAN! PLEASE, THIS IS URGENT! Solve √(k7) = √(k) 1 for k.

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Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0take square both sides to cancel out square root

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i tried that and i got this \[k7= k1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or is it k+1? i think its K1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can anyone help me please?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\sqrt{k7}=\sqrt{k}1 ?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm trying to have it done by the time my best friend gets on.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the choices are: 20 16 4 8

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0oh nvm I didn't notice the parentheses

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you will need this on the right hand side \[(ab)^2=a^2+b^22ab\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sqrt{k7} = \sqrt{k} 1\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2have you expanded the right hand side yet?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont know what that means

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i got to this point\[k7 = k1\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\sqrt{k7}=\sqrt{k}1 \\ \text{ Square both sides } (\sqrt{k7})^2=(\sqrt{k}1)^2 \\ k7=(\sqrt{k}1)^2\] recall \[(ab)^2=a^2+b^22ab\] you can use this to expand the square part on the right hand side

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooooohhhhhhh soooooo\[k+12k\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0right? so \[k+12k=k7\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you had a sqrt( ) on k

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[(ab)^2=a^2+b^22ab \\ (\sqrt{k}1)^2=(\sqrt{k})^2 +(1)^22(\sqrt{k})(1) \\ (\sqrt{k}1)^2=k+12 \sqrt{k}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh right. \[k7 = k+1  2\sqrt{k}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so you really have \[k7=k+12 \sqrt{k} \text{ or subtracting } k \text{ on both sides } 7=12 \sqrt{k}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2isolate the sqrt(k) part

freckles
 one year ago
Best ResponseYou've already chosen the best response.2divide both sides by 2

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yep and you can check that by pluggin it in

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0THANK YOUUU soooooooooo muchhh

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\sqrt{k7}=\sqrt{k}1 \\ k=16 \\ \sqrt{167}=\sqrt{9}=3 \\ \text{ and the other side } \sqrt{16}1=41=3 \] 3=3 k=16 is definitely a solution since it gives a true equation when pluggin it back in
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