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kiara23
Group Title
PLEASE HELP ME Solve 2x^2+3x9=0
a. x=(3+/3i)sqr7/4
b. x=(3+/9i)sqrt7/4
c. x=(3+/3i)sqrt7/4
d. x=3+/9i)sqrt7/4
 2 years ago
 2 years ago
kiara23 Group Title
PLEASE HELP ME Solve 2x^2+3x9=0 a. x=(3+/3i)sqr7/4 b. x=(3+/9i)sqrt7/4 c. x=(3+/3i)sqrt7/4 d. x=3+/9i)sqrt7/4
 2 years ago
 2 years ago

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agentgamer Group TitleBest ResponseYou've already chosen the best response.0
{"x = 3/4 (1i sqrt(7)) ~~ 0.751.98431 i", "x = 3/4 (1+i sqrt(7)) ~~ \ 0.75+1.98431 i"}
 2 years ago

kiara23 Group TitleBest ResponseYou've already chosen the best response.1
Thats not right..
 2 years ago

DHASHNI Group TitleBest ResponseYou've already chosen the best response.1
Divide both sides by 2: x^2(3 x)/2+9/2 = 0 Subtract 9/2 from both sides: x^2(3 x)/2 = 9/2 Add 9/16 to both sides: x^2(3 x)/2+9/16 = 63/16 Factor the left hand side: (x3/4)^2 = 63/16 Take the square root of both sides: sqrt(x3/4)^2 = (3 i sqrt(7))/4 Eliminate the absolute value: x3/4 = (3 i sqrt(7))/4 or x3/4 = (3 i sqrt(7))/4 Add 3/4 to both sides: x = 3/4 (1i sqrt(7)) or x3/4 = (3 i sqrt(7))/4 Add 3/4 to both sides: x = 3/4 (1i sqrt(7)) or x = 3/4 (1+i sqrt(7))
 2 years ago
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