how do I find the quadratic equation with the solutions given - a+bi and a-bi??
K kwanjo New member Joined Nov 6, 2010 Messages 1 Nov 6, 2010 #1 how do I find the quadratic equation with the solutions given - a+bi and a-bi??
L lookagain Elite Member Joined Aug 22, 2010 Messages 3,251 Nov 6, 2010 #2 kwanjo said: how do I find the quadratic equation with the solutions given - a+bi and a-bi?? Click to expand... kwanjo, suppose the variable is \(\displaystyle x.\) \(\displaystyle (x - root_1)(x - root_2) = 0\) The quadratic equation can start off with \(\displaystyle [x - (a + bi)][x - (a - bi)] = 0\) \(\displaystyle x^2 - (a - bi)x - (a + bi)x + (a + bi)(a - bi) = 0\) \(\displaystyle x^2 - ax + bix - ax - bix + a^2 - abi + abi - b^2i^2 = 0\) \(\displaystyle x^2 - 2ax + a^2 - b^2(-1) = 0\) and end with \(\displaystyle \boxed{x^2 - 2ax + a^2 + b^2 = 0}\) Note: It looks as if you have a hyphen, then a space, and then the first root of \(\displaystyle "a + bi."\) That's what I based my work on.
kwanjo said: how do I find the quadratic equation with the solutions given - a+bi and a-bi?? Click to expand... kwanjo, suppose the variable is \(\displaystyle x.\) \(\displaystyle (x - root_1)(x - root_2) = 0\) The quadratic equation can start off with \(\displaystyle [x - (a + bi)][x - (a - bi)] = 0\) \(\displaystyle x^2 - (a - bi)x - (a + bi)x + (a + bi)(a - bi) = 0\) \(\displaystyle x^2 - ax + bix - ax - bix + a^2 - abi + abi - b^2i^2 = 0\) \(\displaystyle x^2 - 2ax + a^2 - b^2(-1) = 0\) and end with \(\displaystyle \boxed{x^2 - 2ax + a^2 + b^2 = 0}\) Note: It looks as if you have a hyphen, then a space, and then the first root of \(\displaystyle "a + bi."\) That's what I based my work on.
