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1) What is the smallest distance between the origin and a point on the graph of $y=\dfrac{1}{\sqrt{2}}\left(x^2-3\right)$?

For this problem, I have tried to use the distance formula and simplify and get a quadratic, but something is messed up... I have no idea what I did wrong...

2) Let $a,b$ be real numbers, and let $x_1,x_2$ be the roots of the quadratic equation $x^2+ax+b=0$.

Prove that if $x_1,x_2$ are real and nonzero, $\frac 1{x_1}+\frac 1{x_2}<1$, and $b>0$, then $|a+2|>2$.

My proof so far : Proof

How would I do the second part of this proof?

1 Answers 1

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Hints:

  • The square of the distance between a point on the curve and the origin is:

$$x^2+y^2=x^2+\frac{1}{2}\left(x^2-3\right)^2=\frac{1}{2}\left(x^4 - 4x^2+9\right) = \frac{1}{2}\left( \left(x^2-2\right)^2 +5\right)$$

  • From the given relation, and the condition that the discriminant be non-negative: $$ \begin{cases} \begin{align} a & \gt -b \\ a^2 & \ge 4b \end{align} \end{cases} $$ Multiplying the first inequality by $4$ and adding together:

$$ a^2 + 4a \gt 4b - 4b = 0 \;\;\iff\;\; a^2+4a+4 \gt 4 \;\;\iff\;\; (a+2)^2 \gt 4 $$