One root of the quadratic equation y^2 + by + c = 0, where b and c are real constants, is 3 - isqrt3. What are b and c?

Blizzardshine Jul 1, 2022

#1**0 **

Note that \(3 - i\sqrt3 = 3 - \sqrt{-3}\).

Now, recall the quadratic formula: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\). Substituting \(a = 1\), we have \(3 - \sqrt {-3} = {-b - \sqrt{b^2-4c} \over 2}\) (take the minus root, not the plus)

Multiplying both sides by 2 gives us: \(6 - 2 \sqrt {-3} = -b - \sqrt{ b^2 - 4ac}\). Simplify the left-hand side: \(6 - \sqrt{-12} = -b - \sqrt {b^2 - 4ac}\).

Note that \(6 = -b\), meaning \(b = -6\). Plugging this into the discriminant, we have \({36 - 4c} = -12\) (disregard the square root because both have them)

Solving, we find that \(c = 12\).

BuilderBoi Jul 1, 2022

#2**+1 **

Can you make your answer a little more detailed? Like, why did you make a=1, why only take the minus root, how is 6=-b, and how did you get 36-4c=-12?

Blizzardshine
Jul 1, 2022