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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?

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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

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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
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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