Squircles!

Algebra Level 4

A squircle is a mathematical shape with properties between those of a square and those of a circle. A squircle centered at the origin is defined by the equation: $x^4+y^4 = r^4$

What is the sum of the squares of the distinct $x$ -coordinates of the intersections of the graphs: $x^4+y^4 = 8$ and $y = \sqrt[4]{-6x^2+16}$

The answer is 4.

1 solution

Sharad Roy
May 2, 2014

$\quad \quad \quad \quad \quad \quad { \quad \quad x }^{ 4 }+{ y }^{ 4 }\quad =\quad 8\quad and\quad y\quad =\quad \sqrt [ 4 ]{ -6{ x }^{ 2\quad }+\quad 16 } \\ Substituting\quad the\quad value\quad of\quad 'y'\quad in\quad first\quad equation,\quad we\quad get:\\ \quad \quad \quad \quad \quad \quad \quad \quad { x }^{ 4 }\quad +\quad (-{ 6x }^{ 2 }\quad +\quad 16)\quad -\quad 8\quad =\quad 0\\ \quad \quad \quad \quad \quad \quad \Rightarrow \quad { x }^{ 4 }\quad -\quad 6{ x }^{ 2 }\quad +\quad 8\quad =\quad 0\\ \quad \quad \quad \quad \quad \quad \Rightarrow \quad ({ x }^{ 2 }-\quad 4)({ x }^{ 2 }-\quad 2)\quad =\quad 0\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \therefore \quad { x }^{ 2 }\quad =\quad 2,4\\ But\quad when\quad we\quad substitute\quad the\quad value\quad of\quad { x }^{ 2 }\quad =\quad 4\quad in\quad second\quad \\ equation,\quad we\quad get\quad -ve\quad no.\quad inside\quad it\quad and\quad then\quad the\quad \\ value\quad of\quad y\quad becomes\quad imaginary.\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \therefore \quad { x }^{ 2 }\quad =\quad 2\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \Rightarrow \quad x\quad =\quad \pm \sqrt { 2 } \\ Therefore,\quad sum\quad of\quad square\quad of\quad x-coordinates\quad =\quad 2\quad +\quad 2\quad =\quad 4$

Squircles!

The answer is 4.

1 solution

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