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 $$\dfrac{bc}{a} $$

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 $$\dfrac{bc}{a} $$

Uniteasy Admin · Accepted Answer

From the given conditions, the following equation are obtained
$$a+(-2b)=\sqrt{2}$$n
and
$$a(-2b) = \sqrt{3}c^2+\dfrac{1}{2} $$n
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 $$0$$.

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