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An analysis of a circuit shown in Fig yields the quadratic equation for the current I as 3 I 2 + 6 I = 45,
An analysis of a circuit shown in Fig yields the quadratic equation for the current I as 3 I 2 + 6 I = 45, where I is in amps.
(a) Rewrite the above equation in the form of I 2 + a I + b = 0, where a and b are constants.
(b) Solve the equation in par (a) by each of the following methods: factoring, completing the square, and the quadratic formula
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a 3I 2 6I 45 3I 2 6I 450 I 2 63I 4530 I 2 2I 15 ...
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