How do you solve the system $x^2+y^2=16$ and $x^2-5y=5$ ?

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Answer 1 · 2016-07-29T21:29:23

$(+-sqrt(-15/2+(5sqrt(69))/2), -5/2+sqrt(69)/2)$

$(+-(sqrt(15/2+(5sqrt(69))/2))i, -5/2-sqrt(69)/2)$

Subtract the second equation from the first and rearrange to get:

$y^2+5y-11 = 0$

$y = (-5+-sqrt(5^2-4(1)(-11)))/2$

$=(-5+-sqrt(25+44))/2$

$=(-5+-sqrt(69))/2$

Then from the second equation, we have:

$x^2=5+5y = 5+5((-5+-sqrt(69))/2) = -15/2+-(5sqrt(69))/2$

Hence:

$x = +-sqrt(-15/2+-(5sqrt(69))/2)$

Note that $sqrt(69) > sqrt(64) = 8$ , hence $(5sqrt(69))/2 > 15/2$

So we find $2$ Real values of $x$ and $2$ non-Real Complex values of $x$ .

So the solutions are:

$(+-(sqrt(-15/2+(5sqrt(69))/2)), -5/2+sqrt(69)/2)$

$(+-(sqrt(15/2+(5sqrt(69))/2))i, -5/2-sqrt(69)/2)$

How do you solve the system x^2+y^2=16x^2+y^2=16 and x^2-5y=5x^2-5y=5?