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happykiddo
 3 years ago
solve the equation by finding the square root
x^29=0
i got x=4 + or  √4
happykiddo
 3 years ago
solve the equation by finding the square root x^29=0 i got x=4 + or  √4

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JuanitaM
 3 years ago
Best ResponseYou've already chosen the best response.0i think you meant to type 3

JuanitaM
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360110109271:dw to solve by square root; get the x^2 term by itself and take the square root of each side.

happykiddo
 3 years ago
Best ResponseYou've already chosen the best response.0i posted the answer for the wrong question. sorry. would you be willing to help again heres the my actual question 2x^264=0 and i got x=4 + or  √4 for this question

suivpasyonw
 3 years ago
Best ResponseYou've already chosen the best response.1\[2x ^{2}=64\] \[x ^{2}=32\] \[x=+\sqrt{32}\]

happykiddo
 3 years ago
Best ResponseYou've already chosen the best response.0could i simplify √32 ?

happykiddo
 3 years ago
Best ResponseYou've already chosen the best response.0like x=6+ √27 can be simplified into x=6+ 3√3

suivpasyonw
 3 years ago
Best ResponseYou've already chosen the best response.1of course it would be \[\sqrt{16\times2}\] so \[\sqrt{16}\times \sqrt{2}\]=\[4\sqrt{2}\]
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