Thread: help with equation

{y^4+x^2-4=0
{2y^2-x^2+1=0

i need to find how many solutions can be, and to mark them on a graph.

and then to know the zone of the shape that created as a result from the
mark.

hope u can understand my question.
and yes my English is not so good.
thank you all.

Originally Posted by ZOOZ

{y^4+x^2-4=0
{2y^2-x^2+1=0

i need to find how many solutions can be, and to mark them on a graph.

and then to know the zone of the shape that created as a result from the
mark.

hope u can understand my question.
and yes my English is not so good.
thank you all.

From the first equation, you see that $\displaystyle \displaystyle x^2 = 4 - y^4$. Replace $\displaystyle \displaystyle x^2$ in the second equation with this expression. you get

$\displaystyle \displaystyle 2y^2 - (4 - y^4) + 1 = 0$

Now solve that equation for $\displaystyle \displaystyle y$. Then plug each of those y-values into the first equation to get the corresponding x-values.

Can you finish? you should have 4 points as your answer.

well thanks for ur hint but i don't know how to keep form this point 2y^2-3+y^4

what i am doing with the y^4?

hope u can help me

Originally Posted by ZOOZ

well thanks for ur hint but i don't know how to keep form this point 2y^2-3+y^4

what i am doing with the y^4?

hope u can help me

Think of it as a quadratic equation:

$\displaystyle \displaystyle y^4 + 2y^2 - 3 = 0$

Is the same as saying

$\displaystyle \displaystyle (y^2)^2 + 2(y^2) - 3 = 0$

which is a quadratic equation in $\displaystyle \displaystyle y^2$. If you have trouble seeing this, let $\displaystyle \displaystyle a = y^2$, then the equation becomes

$\displaystyle \displaystyle a^2 + 2a - 3 = 0$

Solve that for $\displaystyle \displaystyle a$, then replace the $\displaystyle \displaystyle a$ with $\displaystyle \displaystyle y^2$ and solve for $\displaystyle \displaystyle y$

hi i did as u told me and i got X1=1 X2=-3

but i know i need to get 4 point so what i need to do?

thanks for ur help.

Originally Posted by ZOOZ

hi i did as u told me and i got X1=1 X2=-3

but i know i need to get 4 point so what i need to do?

thanks for ur help.

what? what are x1 and x2??

i think you mean you got $\displaystyle \displaystyle y^2 = 1$ or $\displaystyle \displaystyle y^2 = -3$. Clearly the last one is not possible for real numbers, so just take the first one. Then that means $\displaystyle \displaystyle y = \pm 1$. Plug each of these values into the first equation and solve for $\displaystyle \displaystyle x$. you will get 2 x-values each, giving you a total of four points.

well ok if i take the value 1 from the y=+-1
and put him in the first equation is will give me X^2=3

let me know if u mean to take it and put it instead of y^4 ?

appreciate ur patience

Originally Posted by ZOOZ

well ok if i take the value 1 from the y=+-1
and put him in the first equation is will give me X^2=3

let me know if u mean to take it and put it instead of y^4 ?

appreciate ur patience

if $\displaystyle \displaystyle x^2 = 3$, then what is $\displaystyle \displaystyle x$?

hi thanks so much i success to solve it
so really thank u.