Find the minimum value of $|x| - |y|$

Algebra Level 4

$\log_4 (x+2y) + \log_4 (x-2y) = 1.$ Find the minimum value of $|x| - |y|$ .

\sqrt{2}

\sqrt{3}

\sqrt[3]{3}

\sqrt[3]{2}

1 solution

Trevor Arashiro
Feb 17, 2016

$x^2-4y^2=4$

Let $|x|-|y|=n\longrightarrow |x|-n=|y|$

We can't plug this in for y, however, we can plug it into $y^2$

$x^2-4(|x|-n)^2=4$

$0=3x^2-8|x|n+4n^2+4$

Quadratic

$|x|=\dfrac{8n\pm \sqrt{16n^2-48}}{6}$

$|x|\epsilon \Bbb{R^+}$ . Thus the discriminant must be $\geq0$ So $\boxed{n\geq \sqrt{3}}$

overrated :p

Samanvay Vajpayee - 5 years, 2 months ago

@Trevor Arashiro This is nice!! I used trigonometric substitution....... y = tan(theta) and x = 2sec(theta)

Aaghaz Mahajan - 3 years ago