Question

WILL MEDAL AND FAN! PLEASE, THIS IS URGENT!

Solve √(k-7) = √(k) -1 for k.

Answer 1

take square both sides to cancel out square root

Answer 2

i tried that
and i got this \[k-7= k-1\]

Answer 3

or is it k+1? i think its K-1

Answer 4

can anyone help me please?

Answer 5

\[\sqrt{k-7}=\sqrt{k}-1 ?\]

Answer 6

I'm trying to have it done by the time my best friend gets on.

Answer 7

yep!

Answer 8

the choices are:
20
16
4
8

Answer 9

oh nvm I didn't notice the parentheses

Answer 10

square both sides

Answer 11

you will need this on the right hand side
\[(a-b)^2=a^2+b^2-2ab\]

Answer 12

\[\sqrt{k-7} = \sqrt{k} -1\]

Answer 13

have you expanded the right hand side yet?

Answer 14

i dont know what that means

Answer 15

i got to this point\[k-7 = k-1\]

Answer 16

\[\sqrt{k-7}=\sqrt{k}-1 \\ \text{ Square both sides } (\sqrt{k-7})^2=(\sqrt{k}-1)^2 \\ k-7=(\sqrt{k}-1)^2\]
recall
\[(a-b)^2=a^2+b^2-2ab\]
you can use this to expand the square part on the right hand side

Answer 17

ooooohhhhhhh soooooo\[k+1-2k\]

Answer 18

right?
so \[k+1-2k=k-7\]

Answer 19

close

Answer 20

you had a sqrt( ) on k

Answer 21

\[(a-b)^2=a^2+b^2-2ab \\ (\sqrt{k}-1)^2=(\sqrt{k})^2 +(1)^2-2(\sqrt{k})(1) \\ (\sqrt{k}-1)^2=k+1-2 \sqrt{k}\]

Answer 22

oh right. \[k-7 = k+1 - 2\sqrt{k}\]

Answer 23

so you really have
\[k-7=k+1-2 \sqrt{k} \text{ or subtracting } k \text{ on both sides } -7=1-2 \sqrt{k}\]

Answer 24

isolate the sqrt(k) part

Answer 25

k\[-2\sqrt{k}=-8\]

Answer 26

right?

Answer 27

yeah

Answer 28

divide both sides by -2

Answer 29

\[\sqrt{k}=4\]

Answer 30

k=16!

Answer 31

yep
and you can check that by pluggin it in

Answer 32

THANK YOUUU soooooooooo muchhh

Answer 33

Im a fan a medals

Answer 34

\[\sqrt{k-7}=\sqrt{k}-1 \\ k=16 \\ \sqrt{16-7}=\sqrt{9}=3 \\ \text{ and the other side } \sqrt{16}-1=4-1=3 \]
3=3
k=16 is definitely a solution since it gives a true equation when pluggin it back in

Answer 35

thank youuu

Answer 36

np