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Destinyyyy
 one year ago
Dont remember how to do this one...
Destinyyyy
 one year ago
Dont remember how to do this one...

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Destinyyyy
 one year ago
Best ResponseYou've already chosen the best response.1Solve the equation by the square root property. (x9)^2=9

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1square root property so you should take square root both sides to cancel out the square

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1square root and square cancel each other out bec you can convert square root to 1/2 exponent \[\huge\rm\sqrt{x}=x^\frac{ 1 }{ 2 }\] so when you take square \[(x^\frac{ 1 }{ 2 })^2\] both 2 cancels out then you have left with just x so that's why we should take square root to cancel out square and square to cancel out the square root

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1so take square root both sides what did you get ?

Destinyyyy
 one year ago
Best ResponseYou've already chosen the best response.1Sorry but I dont understand

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1thats fine so what u didn't understand ?

Destinyyyy
 one year ago
Best ResponseYou've already chosen the best response.1Um basically everything you said :/

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1otay so take square root both sides to cancel out the square \[\huge\rm \sqrt{(x9)^2}=\sqrt{9}\]

Destinyyyy
 one year ago
Best ResponseYou've already chosen the best response.1That turns into x9= + 3

Destinyyyy
 one year ago
Best ResponseYou've already chosen the best response.1The solution set is 6 and 12

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\[\huge\rm x9 = \pm 3\] so x9 = 3 and x9 =3

Destinyyyy
 one year ago
Best ResponseYou've already chosen the best response.1Wow I didnt remember that at all till you showed me with the square root. Lol thank you again

Destinyyyy
 one year ago
Best ResponseYou've already chosen the best response.1I might be asking for more of your help later. Im on question 4 out of 23 .
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