Finding the Square Root of a Quadratic Function
Date: 10/30/2002 at 15:09:12 From: Ryan MacRae Subject: Finding the square root of a Quadratic Function. I was receintly assinged the problem Find the square root of 3+4i I am in advanced college math and am totally stumped by this question. Please help me... Thank you for your time. Ryan
Date: 10/30/2002 at 17:46:55 From: Doctor Paul Subject: Re: Finding the square root of a Quadratic Function. If sqrt(3+4*i) = a + b*i where a and b are real numbers, then squaring both sides gives: (a + b*i) * (a + b*i) = 3 + 4*i a^2 + 2*a*b*i - b^2 = 3 + 4*i equating coefficients gives: a^2 - b^2 = 3 and 2*a*b = 4 Solve for b in the second equation and substitute into the first equation: b = 2/a so a^2 - (2/a)^2 = 3 a^2 - 4/a^2 = 3 multiply both sides by a^2: a^4 - 3*a^2 - 4 = 0 This is quadratic in a^2 so we make a substitution: Now let x = a^2 So we have: x^2 - 3*x - 4 = 0 (x-4)*(x+1) = 0 So x = 4 or x = -1 This gives a = 2, -2, i, -i We said above that a had to be real, so it must be the case that a = 2 or a = -2. Now, if a = 2 then b = 1 and if a = -2 then b = -1 So we have: sqrt(3 + 4*i) = (2 + i) or -(2 + i) I hope this helps. Please write back if you'd like to talk about this some more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/
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