- #1

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## Homework Statement

Solve for x and y:

x

^{2}+ (√8)x*sin[(√2)xy] +2 = 0

## Homework Equations

## The Attempt at a Solution

Other than decomposing the root 8 i don't know what else to do. any hints? thanks.

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- Thread starter kscplay
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- #1

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Solve for x and y:

x

Other than decomposing the root 8 i don't know what else to do. any hints? thanks.

- #2

ehild

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Hi kscplay,

Transform the equation into the form (x+a)^{2}+b=0

ehild

Transform the equation into the form (x+a)

ehild

- #3

Mark44

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## Homework Statement

Solve for x and y:

x^{2}+ (√8)x*sin[(√2)xy] +2 = 0

QUOTE]

ehild, I don't see how this will lead anywhere, due to the presence of x in the sine factor.Hi kscplay,

Transform the equation into the form (x+a)^{2}+b=0

- #4

ehild

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The left hand side can not be negative. The right hand side ?...

ehild

- #5

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this might be a silly question, but wouldn't you first isolate y?

- #6

ehild

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this might be a silly question, but wouldn't you first isolate y?

That could be also a way to find the solution. But I would look at the right hand side of my equation and see if it can be positive.

ehild

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- #7

ehild

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this might be a silly question, but wouldn't you first isolate y?

If we isolate y we get [tex]y=\frac{1}{\sqrt{2}x}\arcsin(-\frac{x^2+2}{2\sqrt{2}x})[/tex] The argument of arcsin has to be in the range [-1,1]. What does it mean for x?

My method of transforming the equation to [itex](x+\sqrt{2}\sin(\sqrt{2}xy)^2=-2+2sin^2(\sqrt{2}xy)[/itex] shows that the left hand side can not be negative, the right-hand side can not be positive so both have to be zero.

[tex]x+\sqrt{2}\sin(\sqrt{2}xy)=0[/tex]

[tex]\sin^2(\sqrt{2}xy)=1 [/tex]

two equations, two unknowns...

ehild

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- #8

Mentallic

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If we isolate y we get [tex]y=\frac{1}{\sqrt{2}x}\arcsin(-\frac{x^2+2}{2\sqrt{2}x})[/tex] The argument of arcsin has to be in the range [-1,1]. What does it mean for x?

My method of transforming the equation to (x+√2sin[√2xy])^2=-2+2sin^2[√2xy]) shows that the left hand side can not be negative, the right-hand side can not be positive so both have to be zero.

[tex]x+\sqrt{2}\sin^2(\sqrt{2}xy)=0[/tex]

[tex]\sin^2(\sqrt{2}xy)=1 [/tex]

two equations, two unknowns...

ehild

That's genius! I'm glad I didn't miss out on reading this

- #9

ehild

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There was a typo , I corrected it in post #7. The square should not be there in my first equation. Correctly it is [tex]x+\sqrt{2}\sin(\sqrt{2}xy)=0[/tex]

ehild

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- #10

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That's a smart idea. Thanks ehild :)

- #11

ehild

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You are welcome, kscplay.

The other way, jkristia suggested would work too.

[tex]y=-\frac{1}{\sqrt{2}x}\arcsin \left( \frac{x^2+2}{2\sqrt{2}x} \right)[/tex]

The magnitude of the argument of the arcsin function can not exceed 1:

[tex] \left| \frac{x^2+2}{2 \sqrt{2} x} \right| \leq 1[/tex]

The function

[tex]

\frac{x^2+2}{2\sqrt{2}x}[/tex] has extrema at x=±√2, minimum (1) for positive x and maximum (-1) for negative x. Only these values are allowed for x.

ehild

The other way, jkristia suggested would work too.

[tex]y=-\frac{1}{\sqrt{2}x}\arcsin \left( \frac{x^2+2}{2\sqrt{2}x} \right)[/tex]

The magnitude of the argument of the arcsin function can not exceed 1:

[tex] \left| \frac{x^2+2}{2 \sqrt{2} x} \right| \leq 1[/tex]

The function

[tex]

\frac{x^2+2}{2\sqrt{2}x}[/tex] has extrema at x=±√2, minimum (1) for positive x and maximum (-1) for negative x. Only these values are allowed for x.

ehild

Last edited:

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