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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?

 Jul 1, 2022
 #1
avatar+2448 
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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 \(c = 12\).

 Jul 1, 2022
 #2
avatar+236 
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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

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