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Solve the equation 36x^2+25=0 a. +6/5i b. +5/6i c. +25/36i d. +36/25i
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\(\bf \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\qquad thus \\ \quad \\ 36x^2+25=0\quad \begin{cases} 36\to 6^2\\ 25\to 5^2 \end{cases}\qquad \implies 6^2x^2+5^2=0\implies (6x)^2+5^2=0\)
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