Question 985922
{{{y=x^2-10}}}
{{{9x^2-y^2=90}}}
----------------

{{{y=x^2-10}}}
{{{9x^2-90=y^2}}}
----------------

{{{y=x^2-10}}}
{{{y=sqrt(9x^2-90)}}}
----------------

{{{y=x^2-10}}}
{{{y=sqrt(9(x^2-10))}}}
----------------

{{{y=x^2-10}}}
{{{y=3sqrt(x^2-10)}}}
----------------

if left sides same, then we have

{{{x^2-10=3sqrt(x^2-10)}}}

{{{x^2/3-10/3=sqrt(x^2-10)}}}............square both sides

{{{(x^2/3-10/3)^2=(sqrt(x^2-10))^2}}}

{{{(x^2/3)^2-2(x^2/3)*(-10/3)+(-10/3)^2=x^2-10}}}

{{{(x^4/9)-(20x^2/9)+(100/9)=x^2-10}}}.....both sides multiply by {{{9}}}

{{{9(x^4/9)-9(20x^2/9)+9(100/9)=9x^2-90}}} 

{{{x^4-20x^2+100=9x^2-90}}} 

{{{x^4-20x^2+100-9x^2+90=0}}}

{{{x^4-29x^2+190=0}}}....factor

{{{x^4-19x^2-10x^2+190=0}}}

{{{(x^4-19x^2)-(10x^2-190)=0}}}

{{{x^2(x^2-19)-10(x^2-19)=0}}}

{{{(x^2-19)(x^2-10) = 0}}}

zeros:


if {{{x^2-19 = 0}}}=>{{{x^2=19 }}}=>{{{x=sqrt(19) }}} or {{{x=-sqrt(19) }}}
if {{{x^2-10 = 0}}}=>{{{x^2=10 }}}=>{{{x=sqrt(10) }}} or {{{x=-sqrt(10) }}}

approximately:

{{{x=4.4 }}} or {{{x=-4.4 }}}
{{{x=3.2 }}} or {{{x=-3.2 }}}

{{{y=x^2-10}}}
{{{9x^2-y^2=90}}}

for {{{x = sqrt(19)}}}=> {{{y=(sqrt(19))^2-10}}}=> {{{y=19-10}}}=>{{{y = 9}}}
for {{{x = sqrt(10)}}}=> {{{y=(sqrt(10))^2-10}}}=> {{{y=10-10}}}=>{{{y = 0}}}

for {{{x = sqrt(19)}}}=>{{{9(sqrt(19))^2-90=y^2}}}=>{{{171-90=y^2}}}=>{{{81=y^2}}}=>{{{y = 9}}} and =>{{{y = -9}}}

for {{{x = sqrt(10)}}}=>{{{9(sqrt(10))^2-90=y^2}}}=>{{{90-90=y^2}}}=>{{{9=y^2}}}=>{{{y = 0}}} 

check:

{{{drawing( 600, 600, -10, 10, -20, 10,
circle(sqrt(10),0,.12),locate(3.2,0.9,p(sqrt(10),0)),
circle(-sqrt(10),0,.12),locate(-3.2,0.9,p(-sqrt(10),0)),
circle(-sqrt(19),9,.12),locate(-sqrt(19),9,p(-sqrt(19),9)),
circle(sqrt(19),9,.12),locate(sqrt(19),9,p(sqrt(19),9)),

 graph( 600, 600, -10, 10, -20, 10,x^2-10,sqrt(9(x^2-10)),-sqrt(9(x^2-10)))) }}}