WILL MEDAL AND FAN! PLEASE, THIS IS URGENT! Solve √(k-7) = √(k) -1 for k.

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WILL MEDAL AND FAN! PLEASE, THIS IS URGENT! Solve √(k-7) = √(k) -1 for k.

Mathematics
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take square both sides to cancel out square root
i tried that and i got this \[k-7= k-1\]
or is it k+1? i think its K-1

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can anyone help me please?
\[\sqrt{k-7}=\sqrt{k}-1 ?\]
I'm trying to have it done by the time my best friend gets on.
yep!
the choices are: 20 16 4 8
oh nvm I didn't notice the parentheses
square both sides
you will need this on the right hand side \[(a-b)^2=a^2+b^2-2ab\]
\[\sqrt{k-7} = \sqrt{k} -1\]
have you expanded the right hand side yet?
i dont know what that means
i got to this point\[k-7 = k-1\]
\[\sqrt{k-7}=\sqrt{k}-1 \\ \text{ Square both sides } (\sqrt{k-7})^2=(\sqrt{k}-1)^2 \\ k-7=(\sqrt{k}-1)^2\] recall \[(a-b)^2=a^2+b^2-2ab\] you can use this to expand the square part on the right hand side
ooooohhhhhhh soooooo\[k+1-2k\]
right? so \[k+1-2k=k-7\]
close
you had a sqrt( ) on k
\[(a-b)^2=a^2+b^2-2ab \\ (\sqrt{k}-1)^2=(\sqrt{k})^2 +(1)^2-2(\sqrt{k})(1) \\ (\sqrt{k}-1)^2=k+1-2 \sqrt{k}\]
oh right. \[k-7 = k+1 - 2\sqrt{k}\]
so you really have \[k-7=k+1-2 \sqrt{k} \text{ or subtracting } k \text{ on both sides } -7=1-2 \sqrt{k}\]
isolate the sqrt(k) part
k\[-2\sqrt{k}=-8\]
right?
yeah
divide both sides by -2
\[\sqrt{k}=4\]
k=16!
yep and you can check that by pluggin it in
THANK YOUUU soooooooooo muchhh
Im a fan a medals
\[\sqrt{k-7}=\sqrt{k}-1 \\ k=16 \\ \sqrt{16-7}=\sqrt{9}=3 \\ \text{ and the other side } \sqrt{16}-1=4-1=3 \] 3=3 k=16 is definitely a solution since it gives a true equation when pluggin it back in
thank youuu
np

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