Quadratic Genius

Algebra Level 4

$If\quad x\quad and\quad y\quad are\quad width\quad and\quad hight\quad of\quad a\quad \\ rectangle\quad respectively,\quad Find\quad the\quad area?\\ **\quad Use\quad real\quad roots\quad **\quad \\ \\ \\ x=\sqrt { \left( 73+2{ y }^{ 2 } \right) \left( { y }^{ -2 }-\frac { 5 }{ 36 } \right) } \\ \\ y={ \left( \frac { \frac { 15 }{ 24 } { x }^{ -2 } }{ 1-\frac { 3 }{ 2 } { x }^{ -1 } } +{ x }^{ -2 } \right) }^{ \frac { -1 }{ 2 } }\\ \\ \\ \sqrt { y-x } =\quad i$

The answer is 6.

1 solution

Rishabh Jain
Dec 23, 2015

Squaring the equation √y-x=i gives x=y+1 substituting this value of x in the second equation $y=(15/24x^-2..........)^-1/2$ we get $5y^3-11y^2+4=0$ gives the solution y=2 (rejecting other irrational values as they don't satisfy the first equation simultaneously) and x=y+1=3. Area=3*2=6

Wait a minute, isn't it going to become x=y+i instead of 1 in place of i?

jake roosenbloom - 5 years, 5 months ago

i^2=-1 that's why y-x=i^2=-1 or x=y+1

Rishabh Jain - 5 years, 5 months ago

Sorry if it sounds stupid, but how is i^2=-1?

jake roosenbloom - 5 years, 5 months ago

@Jake Roosenbloom – i is defined as $√-1$ , squaring we get $i^2=(√-1)^2=-1$ . In fact for any n∈Z $i^(4n+2) =-1$

Rishabh Jain - 5 years, 5 months ago

Quadratic Genius

The answer is 6.

1 solution

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