help with equation

**ZOOZ** · Oct 24th 2010, 09:04 AM

{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.

**Jhevon** · Oct 24th 2010, 09:32 AM

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

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

Now solve that equation for $\displaystyle y$ . 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.

**ZOOZ** · Oct 24th 2010, 09:55 AM

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

**Jhevon** · Oct 24th 2010, 09:58 AM

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:

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

Is the same as saying

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

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

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

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

**ZOOZ** · Oct 24th 2010, 10:44 AM

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.

**Jhevon** · Oct 24th 2010, 10:47 AM

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

**ZOOZ** · Oct 24th 2010, 11:00 AM

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

**Jhevon** · Oct 24th 2010, 11:27 AM

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

**ZOOZ** · Oct 25th 2010, 12:19 AM

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

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