- #1
kscplay
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Homework Statement
Solve for x and y:
x2 + (√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.
kscplay said:Homework Statement
Solve for x and y:
x2 + (√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.ehild said:Hi kscplay,
Transform the equation into the form (x+a)2+b=0
jkristia said:this might be a silly question, but wouldn't you first isolate y?
jkristia said:this might be a silly question, but wouldn't you first isolate y?
ehild said: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
The equation is trying to solve for the values of x and y that satisfy the given equation.
The steps to solve the equation are as follows:
1. Move all constants to one side of the equation and all variables to the other side.
2. Use the quadratic formula to solve for x.
3. Substitute the value of x into the equation and solve for y.
4. Check the solutions by plugging them back into the original equation.
Yes, the equation can be solved algebraically by following the steps mentioned in the previous question.
Yes, there may be restrictions on the values of x and y depending on the given equation. In this equation, the values of x and y must make the expression under the square root positive, otherwise the equation will not have real solutions.
Yes, the equation can have more than one solution. In fact, quadratic equations can have up to two real solutions.