Question 145269
<pre><font size = 4 color = "indigo"><b>
{{{y=2x^2-1}}}
{{{x^2+2y^2=16}}}

First I'll approach it graphically to see what
to expect; then I'll do it algebraically

The graph of {{{y=2x^2-1}}} is this parabola:

{{{graph(200,200,-5,5,-5,5, 2x^2-1) }}}

The graph of {{{x^2+2y^2=16}}} is this ellipse:

{{{drawing(200,200,-5,5,-5,5,
graph(200,200,-5,5,-5,5, sqrt( (16-x^2)/2      ) ),
graph(200,200,-5,5,-5,5, -sqrt( (16-x^2)/2)  ))

 }}}

Put them together on one graph:

The graph of {{{x^2+2y^2=16}}} is this ellipse:

{{{drawing(200,200,-5,5,-5,5, graph(200,200,-5,5,-5,5, 2x^2-1),
graph(200,200,-5,5,-5,5, sqrt( (16-x^2)/2      ) ),
graph(200,200,-5,5,-5,5, -sqrt( (16-x^2)/2)  ))

 }}}

Since these graphs cross in two points, these will be two
real solutions, and since parabolas and circles could cross
as many as 4 times, there will two imaginary solutions as well

{{{y=2x^2-1}}}
{{{x^2+2y^2=16}}}

The first equation is already solved for {{{y}}}, so we substitute
{{{(2x^2-1)}}} for {{{y}}} in the second equation:

{{{x^2+2y^2=16}}}
{{{x^2+2(2x^2-1)^2=16}}}
{{{x^2+2(2x^2-1)(2x^2-1)=16}}}
{{{x^2+2(4x^4-2x^2-2x^2+1)=16}}}
{{{x^2+2(4x^4-4x^2+1)=16}}}
{{{x^2+8x^4-8x^2+2=16}}}
{{{8x^4-7x^2+2=16}}}
{{{8x^4-7x^2-14=0}}}

That does not factor, so we solve that for {{{x^2}}}
using the quadratic formula:

{{{x^2 = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

where {{{a=8}}}, {{{b=-7}}}, {{{c=-14}}}

{{{x^2 = (-(-7) +- sqrt( (-7)^2-4*(8)*(-14) ))/(2*(8)) }}} 

{{{x^2 = (7 +- sqrt( 49+448 ))/16 }}}

{{{x^2 = (7 +- sqrt( 497 ))/16 }}}

So we have two solutions for {{{x^2}}}

{{{x^2 = (7 + sqrt( 497 ))/16 }}} and {{{x^2 = (7 - sqrt( 497 ))/16 }}}

If we substitute the first value of {{{x^2}}} in {{{y=2x^2-1}}}, we get

{{{y=2((7 + sqrt( 497 ))/16)-1=(7 + sqrt( 497 ))/8-1 = (7 + sqrt( 497 ))/8-8/8 = (7 + sqrt( 497 )-8)/8 = (-1 + sqrt( 497 ))/8}}}  

If we substitute the second value of {{{x^2}}} in {{{y=2x^2-1}}}, we get

{{{y=2((7 - sqrt( 497 ))/16)-1=(7 - sqrt( 497 ))/8-1 = (7 - sqrt( 497 ))/8-8/8 = (7 - sqrt( 497 )-8)/8 = (-1 - sqrt( 497 ))/8}}} 

The first value of {{{x^2}}} gives these two solutions for x:

{{{x = sqrt((7 + sqrt( 497 ))/16)= sqrt(7 + sqrt( 497 ))/4    }}} and {{{x = -sqrt((7 + sqrt( 497 ))/16)= -sqrt(7 + sqrt( 497 ))/4    }}}

The second value of {{{x^2}}}gives these two solutions for x:

{{{x = sqrt((7 - sqrt( 497 ))/16)= sqrt(7 - sqrt( 497 ))/4    }}}, {{{x = -sqrt((7 - sqrt( 497 ))/16)= sqrt(7 - sqrt( 497 ))/4    }}},

but since {{{7-sqrt(497)}}} is a negative number, it's going to be
imaginary, so we factor {{{-1}}} out of it, {{{-1(-7+sqrt(497))}}},
so we have

{{{x = sqrt(-1(-7 + sqrt( 497 )))/4   = sqrt(-1)sqrt((-7 + sqrt( 497 )))/4 =i*sqrt(-7 + sqrt( 497 ))/4 }}}

and of course {{{x=-i*sqrt(-7 + sqrt( 497 ))/4 }}}

So we have four solutions for ({{{x}}},{{{y}}})

({{{ sqrt( 7 + sqrt( 497 ) )/4    }}}, {{{(-1 + sqrt( 497 ))/8}}}) = ({{{1.353086675}}}, {{{2.661687101}}}) 
({{{-sqrt( 7 + sqrt( 497 ) )/4    }}}, {{{(-1 + sqrt( 497 ))/8 }}}) = ({{{-1.353086675}}}, {{{2.661687101}}}) 

These are the points where the curves above cross, and are the two
real solutions.

Here are the two imaginary solutions:

({{{ i*sqrt(-7 + sqrt( 497 ) )/4    }}}, {{{(-1 - sqrt( 497 ))/8 }}}) 
({{{-i*sqrt(-7 + sqrt( 497 ) )/4    }}}, {{{(-1 - sqrt( 497 ))/8 }}})
 
Edwin</pre>