Semi Pythagoras Triplet?

Algebra Level 3

Let there be two real numbers $a,b$ that satisfy the given equation:

$a^2+b^2=548464$

The range of the values that $a$ can take for some value of $b$ that satisfies the equation above is of the form $[-x,x]$ for some $x \geq 0$ .

What is the value of $x$ ?

Inspiration

\sqrt[8]{548464}

0

None of these

\sqrt[4]{548464}

\sqrt{548464}

548464

\sqrt[6]{548464}

1

4 solutions

Kushal Patankar
Feb 26, 2015

Equation can be written as $( \sqrt{548464} sin \theta)^2 +( \sqrt{548464} cos \theta)^2 = 548464$ For some arbitrary theta.

$\Rightarrow a= \sqrt{548464} sin \theta$ $\Rightarrow \text{ Range of a is} \left[ - \sqrt{548464}, \sqrt{548464}\right]$

Prasun Biswas - 6 years, 3 months ago

Use $\backslash(\textrm{\sqrt{548464}}\backslash)$ to produce the $\LaTeX$ output as: $\sqrt{548464}$ .

Also, there are some minor typos in the last two lines. Please edit them accordingly.

Prasun Biswas - 6 years, 3 months ago

Saurabh Agrawal
Sep 15, 2015

This problem can be solved just by taking b as 0(zero), which will give maximum range of a.

Jacob T
Aug 6, 2015

First, solve the equation for a. a= $\sqrt{548464 - b^{2}}$ . Since $b^{2}$ is necessarily positive, the largest value of a is when b is zero, so a is $\sqrt{548464}$ . Similarly, the smallest value is - $\sqrt{548464}$ .

In your first line, it should be $|a|=\sqrt{548464-b^2}$ since $a$ can be negative too.

Prasun Biswas - 5 years, 10 months ago

Incredible Mind
Feb 25, 2015

that is eqn of circle

Semi Pythagoras Triplet?

4 solutions

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