Finding Points of Intersection by Substitution

  • #1

Homework Statement


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

[tex]xy+x-2y+3=0[/tex]
[tex]x^2+4y^2-9=0[/tex]

Homework Equations




The Attempt at a Solution



From the second equation I can solve for y:

[tex]y=\frac{\sqrt{9-x^2}}{2}[/tex]

Plug it into the first equation and simplify...

[tex]\frac{x\sqrt{9-x^2}+2x-2\sqrt{9-x^2}+6}{2}=0[/tex]

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

[tex](x\sqrt{9-x^2}-2\sqrt{9-x^2})^2=(-2x-6)^2[/tex]

simplify and you get

[tex]-x(x^3-4x^2-x+60)=0[/tex]

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?
 

Answers and Replies

  • #2
35,229
7,048

Homework Statement


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

[tex]xy+x-2y+3=0[/tex]
[tex]x^2+4y^2-9=0[/tex]

Homework Equations




The Attempt at a Solution



From the second equation I can solve for y:

[tex]y=\frac{\sqrt{9-x^2}}{2}[/tex]
That should be [itex]\pm[/itex].

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

[tex]\frac{x\sqrt{9-x^2}+2x-2\sqrt{9-x^2}+6}{2}=0[/tex]

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

[tex](x\sqrt{9-x^2}-2\sqrt{9-x^2})^2=(-2x-6)^2[/tex]

simplify and you get

[tex]-x(x^3-4x^2-x+60)=0[/tex]

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?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
966

Homework Statement


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

[tex]xy+x-2y+3=0[/tex]

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

x(y+1)= 2y- 3

Solve that for x and avoid square roots.

[tex]x^2+4y^2-9=0[/tex]

Homework Equations




The Attempt at a Solution



From the second equation I can solve for y:

[tex]y=\frac{\sqrt{9-x^2}}{2}[/tex]

Plug it into the first equation and simplify...

[tex]\frac{x\sqrt{9-x^2}+2x-2\sqrt{9-x^2}+6}{2}=0[/tex]

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

[tex](x\sqrt{9-x^2}-2\sqrt{9-x^2})^2=(-2x-6)^2[/tex]

simplify and you get

[tex]-x(x^3-4x^2-x+60)=0[/tex]

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?
 
  • #4
x(y+1)- 2y+ 3= 0

x(y+1)= 2y- 3

Solve that for x and avoid square roots.

Ok, so.

[tex]x=\frac{2y-3}{y+1}[/tex]

You can sub that into the other equation and get

[tex]y(4y^3-8y^2-y+6)=0[/tex] 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.

[tex]\frac{\pm1,2,3,6}{1,2,4}[/tex]

Is this what you had in mind? It is easier than having to do the test with all the factors of 60...
 
  • #5
35,229
7,048
Yes, that's what both HallsOfIvy and I had in mind.
 
  • #6
eumyang
Homework Helper
1,347
10
You can sub that into the other equation and get

[tex]y(4y^3-8y^2-y+6)=0[/tex]
I'm getting -30 inside the parentheses instead of +6. And a plus 8y2 instead of a minus.


69
 

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