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4x^2+98=0 quadratics

1 Answer

5 votes

Answer:


x=+(7)/(√(2))i, \ -(7)/(√(2))i

Explanation:


\sf \rightarrow 4x^2+98=0


\sf \rightarrow 4x^2= -98


\sf \rightarrow x^2= -98/4 = -49/2


\sf \rightarrow x=\pm\sqrt{-(49)/(2)}

  • √-1 = i


\sf \rightarrow x=\pm(7)/(√(2))i


\sf \rightarrow x=+(7)/(√(2))i, \ -(7)/(√(2))i

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