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3+i" or " -3-i "let " sqrt(8+6i)=a+bilarr" square both sides" rArr8+6i=(a+bi)^2=a^2-b^2+2abi "equating the coefficients of real/imaginary parts" a^2-b^2=8to(1) 2ab=6rArrb=3/ato(2) "substitute " b=3/a" in " (1) rArra^2-9/a^2=8 "multiply through by "a^2 rArra^4-9=8a^2 rArra^4-8a^2-9=0 rArr(a^2-9)(a^2+1)=0 a^2+1rArra^2=-1" has no real solutions" a^2-9rArra^2=9rArra=+-3 "substituting in " (2. Because they are unstable with respect to oxidation C_6h_14(g) + 19/2o_2(g) rarr 6co_2(g) + 7h_2o(g) + delta certainly, this is an exothermic reaction, so exothermic that we can utilize the released energy to do useful work First, find 2 real roots of the transformed equation
Next divide them by a to get the 2 real root of the original equation Therefore, the 2 real roots of f'(z) are A science enthusiast!1,622 students helped 1 answer 1 asked I'm an environmental science undergrad
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