Finding Points of Intersection by Substitution

[themadhatter1]

Homework Statement

Find any points of intersection of the graphs by the method of substitution.

$$xy+x-2y+3=0$$
$$x^2+4y^2-9=0$$

Homework Equations

The Attempt at a Solution

From the second equation I can solve for y:

$$y=\frac{\sqrt{9-x^2}}{2}$$

Plug it into the first equation and simplify...

$$\frac{x\sqrt{9-x^2}+2x-2\sqrt{9-x^2}+6}{2}=0$$

Multiply both sides by 2 to get rid of the fraction and get the radicals over to one side, square both sides

$$(x\sqrt{9-x^2}-2\sqrt{9-x^2})^2=(-2x-6)^2$$

simplify and you get

$$-x(x^3-4x^2-x+60)=0$$

I can find x=0 and that Is correct, but the other real root should be -3. I can graph the polynomial (x³-4x²-x+60) on my calculator and see that it has a root of -3 but how can I do it by hand? Any other way besides the rational roots test?

 

[Mark44]

Homework Statement

Find any points of intersection of the graphs by the method of substitution.

$$xy+x-2y+3=0$$
$$x^2+4y^2-9=0$$

Homework Equations

The Attempt at a Solution

From the second equation I can solve for y:

$$y=\frac{\sqrt{9-x^2}}{2}$$

That should be $\pm$.

Also, it might be simpler to solve for one variable, say y, in the first equation.

Plug it into the first equation and simplify...

$$\frac{x\sqrt{9-x^2}+2x-2\sqrt{9-x^2}+6}{2}=0$$

Multiply both sides by 2 to get rid of the fraction and get the radicals over to one side, square both sides

$$(x\sqrt{9-x^2}-2\sqrt{9-x^2})^2=(-2x-6)^2$$

simplify and you get

$$-x(x^3-4x^2-x+60)=0$$

I can find x=0 and that Is correct, but the other real root should be -3. I can graph the polynomial (x³-4x²-x+60) on my calculator and see that it has a root of -3 but how can I do it by hand? Any other way besides the rational roots test?

 

[HallsofIvy]

Homework Statement

Find any points of intersection of the graphs by the method of substitution.

$$xy+x-2y+3=0$$

x(y+1)- 2y+ 3= 0

x(y+1)= 2y- 3

Solve that for x and avoid square roots.

$$x^2+4y^2-9=0$$

Homework Equations

The Attempt at a Solution

From the second equation I can solve for y:

$$y=\frac{\sqrt{9-x^2}}{2}$$

Plug it into the first equation and simplify...

$$\frac{x\sqrt{9-x^2}+2x-2\sqrt{9-x^2}+6}{2}=0$$

Multiply both sides by 2 to get rid of the fraction and get the radicals over to one side, square both sides

$$(x\sqrt{9-x^2}-2\sqrt{9-x^2})^2=(-2x-6)^2$$

simplify and you get

$$-x(x^3-4x^2-x+60)=0$$

I can find x=0 and that Is correct, but the other real root should be -3. I can graph the polynomial (x³-4x²-x+60) on my calculator and see that it has a root of -3 but how can I do it by hand? Any other way besides the rational roots test?

 

[themadhatter1]

x(y+1)- 2y+ 3= 0

x(y+1)= 2y- 3

Solve that for x and avoid square roots.

Ok, so.

$$x=\frac{2y-3}{y+1}$$

You can sub that into the other equation and get

$$y(4y^3-8y^2-y+6)=0$$ Using rational roots test you can find that the root of (4y^3-8y^2-y+6) is 3/2 and of course y= 0 as well.

$$\frac{\pm1,2,3,6}{1,2,4}$$

Is this what you had in mind? It is easier than having to do the test with all the factors of 60...

 

[Mark44]

Yes, that's what both HallsOfIvy and I had in mind.

 

[eumyang]

You can sub that into the other equation and get

$$y(4y^3-8y^2-y+6)=0$$

I'm getting -30 inside the parentheses instead of +6. And a plus 8y² instead of a minus.69