Question

If a,b,c are real numbers such than a=2b+\sqrt{2} and ab+\dfrac{\sqrt{3}}{ 2}c^2+\dfrac{1}{4}=0 , find the value of \dfrac{bc}{a}

Collected in the board: Vieta's Formula

Steven Zheng posted 2 hours ago

Answer

From the given conditions, the following equation are obtained

a+(-2b)=\sqrt{2}
(1)

and

a(-2b) = \sqrt{3}c^2+\dfrac{1}{2}
(2)

Apply Vieta's formula for a quadratic equation. a and -2b are the two roots of the equation

x^2-\sqrt{2}x+ \sqrt{3}c^2+\dfrac{1}{2}=0

Furthermore,

The equation has real roots if and only if its discriminant is great than 0.

(\sqrt{2} )^2-4(\sqrt{3}c^2+\dfrac{1}{2})

=-4\sqrt{3}c^2 > 0

Therefore, c must be zero to meet the criteria. And the value of \dfrac{bc}{a} is equal to 0.

Steven Zheng posted 2 hours ago

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