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