Solving a system of equations

[twiszted]

I am currently taking grade 12 Advanced Functions in Canada.

Here is the question:
x^2 + y^2 = 1
xy = 0.5

This is the work I have done so far but I'm not sure if it is correct or not so I could really use some help please!
- isolate y
y = 0.5/x

x^2 + (0.5/x)^2 = 1
- multiply by x to create common denominator
x^3 + 0.5^2 = x
x^3 - x + 0.25 = 0

I don't know what to do after I get this far. How do I solve for x and y? Please help, greatly appreciated!

Here are the answers but I don't know how to reach them:
x = y = square root 2/2 or x = y = square root -2/2

 

[mmm4444bot]

Square the ratio inside the parentheses before multiplying both sides of the equation. You should then realize that both sides need to be multiplied by x^2, instead of x, in order to clear the ratio.

You will get a fourth-degree polynomial equation that is quadratic in form. To use the Quadratic Formula, first make substitutions for x^4 and x^2.

Have you seen this type of form before?

Here's an example of such substitutions.

Given: 4x^4 - 3x^2 - 1 = 0

This equation is quadratic in form.

Let u = x^2

Then, squaring both sides, we see that u^2 = x^4

Substitute u^2 for x^4, and substitute u for x^2.

4u^2 - 3u - 1 = 0

The Quadratic Formula gives the following solutions for u:

u = -1/4

or

u = 1

Now, reverse the substitutions, and solve the equations for x.

Since u = x^2, we can use the solutions for u to write:

-1/4 = x^2

or

1 = x^2

The solutions for x are:

x = ±i/2

or

x = ±1

Can you try this method?

 

[BigGlenntheHeavy]

\(\displaystyle x^2+y^2 \ = \ 1\)

\(\displaystyle xy \ = \ \frac{1}{2}, \ \implies \ x \ = \ \frac{1}{2y}\)

\(\displaystyle Subbing, \ we \ get \ \frac{1}{4y^2}+y^2 \ = \ 1 \ \implies \ 1+4y^4 \ = \ 4y^2, \ y \ \ne \ 0, \ why?\)

\(\displaystyle Quadratic: \ 4y^4-4y^2+1 \ = \ 0 \ \implies \ y^2 \ = \ \frac{1}{2}, \ hence \ y \ = \ \pm\frac{1}{\sqrt2}\)

\(\displaystyle Therefore, \ when \ x \ = \ \frac{1}{\sqrt2}, \ y \ = \ the \ same \ and \ when \ x \ = \ -\frac{1}{\sqrt2}, \ y \ = \ the \ same.\)

\(\displaystyle I'll \ leave \ the \ check \ up \ to \ you.\)

 

[Deleted member 4993]

I am currently taking grade 12 Advanced Functions in Canada.

Here is the question:
x^2 + y^2 = 1
xy = 0.5

This is the work I have done so far but I'm not sure if it is correct or not so I could really use some help please!
- isolate y
y = 0.5/x

x^2 + (0.5/x)^2 = 1
- multiply by x to create common denominator
x^3 + 0.5^2 = x
x^3 - x + 0.25 = 0

I don't know what to do after I get this far. How do I solve for x and y? Please help, greatly appreciated!

Here are the answers but I don't know how to reach them:
x = y = square root 2/2 or x = y = square root -2/2

For this problem (with these particular numerical values) - there is an apparent shortcut

x[sup:3h1nozy8]2[/sup:3h1nozy8] + y[sup:3h1nozy8]2[/sup:3h1nozy8] - 2xy = 0 ? (x-y)[sup:3h1nozy8]2[/sup:3h1nozy8] = 0 ? x - y = 0 ? x = y

Then

x[sup:3h1nozy8]2[/sup:3h1nozy8] + y[sup:3h1nozy8]2[/sup:3h1nozy8] = 1 ? x[sup:3h1nozy8]2[/sup:3h1nozy8] + x[sup:3h1nozy8]2[/sup:3h1nozy8] = 1 ? 2x[sup:3h1nozy8]2[/sup:3h1nozy8] = 1 ? x[sup:3h1nozy8]2[/sup:3h1nozy8] = 1/2 ? x = ± 1/?2

So the solutions are (1/?2 , 1/?2) and (-1/?2 , -1/?2)

 

[twiszted]

Re:

Square the ratio inside the parentheses before multiplying both sides of the equation. You should then realize that both sides need to be multiplied by x^2, instead of x, in order to clear the ratio.

You will get a fourth-degree polynomial equation that is quadratic in form. To use the Quadratic Formula, first make substitutions for x^4 and x^2.

Have you seen this type of form before?

Here's an example of such substitutions.

Given: 4x^4 - 3x^2 - 1 = 0

This equation is quadratic in form.

Let u = x^2

Then, squaring both sides, we see that u^2 = x^4

Substitute u^2 for x^4, and substitute u for x^2.

4u^2 - 3u - 1 = 0

The Quadratic Formula gives the following solutions for u:

u = -1/4

or

u = 1

Now, reverse the substitutions, and solve the equations for x.

Since u = x^2, we can use the solutions for u to write:

-1/4 = x^2

or

1 = x^2

The solutions for x are:

x = ±i/2

or

x = ±1

Can you try this method?

I just tried this method, and after letting u = x^2, my equation was:
u^2 - u + 0.25 = 0

I then put it into the quadratic formula, but my answers were 1/2 because in side the square root of the quadratic formula was:
(square root) (-1)^2 - 4(1)(0.25)
= 1-1
= 0

therefore, what I am left with is 1 +/- square root 0 / 2. this means that my answer would be 1/2 although I know the answer is square root 2/2

 

[mmm4444bot]

I just tried this method, and after letting u = x^2, my equation was:

u^2 - u + 0.25 = 0 This is good.

I then put it into the quadratic formula …

… this means that my answer would be 1/2 No, it means that u = 1/2. You're not done, yet.

Now that you know the value of u, reverse the substitution.

Since u = x^2, we have 1/2 = x^2.

Solve for x.

 

[twiszted]

Re:

I just tried this method, and after letting u = x^2, my equation was:

u^2 - u + 0.25 = 0 This is good.

I then put it into the quadratic formula …

… this means that my answer would be 1/2 No, it means that u = 1/2. You're not done, yet.

Now that you know the value of u, reverse the substitution.

Since u = x^2, we have 1/2 = x^2.

Solve for x.

Alright great thanks so much. I came up with:

u = 1/2
u = x^2
x^2 = 1/2
(squareroot) x^2 = (squareroot) 1/2
x = +/- 1/square root 2
x = +/- square root 2/2

the only last question i have is how do i prove that x = y?

 

[twiszted]

I am currently taking grade 12 Advanced Functions in Canada.

Here is the question:
x^2 + y^2 = 1
xy = 0.5

This is the work I have done so far but I'm not sure if it is correct or not so I could really use some help please!
- isolate y
y = 0.5/x

x^2 + (0.5/x)^2 = 1
- multiply by x to create common denominator
x^3 + 0.5^2 = x
x^3 - x + 0.25 = 0

I don't know what to do after I get this far. How do I solve for x and y? Please help, greatly appreciated!

Here are the answers but I don't know how to reach them:
x = y = square root 2/2 or x = y = square root -2/2

For this problem (with these particular numerical values) - there is an apparent shortcut

x[sup:1ghphnax]2[/sup:1ghphnax] + y[sup:1ghphnax]2[/sup:1ghphnax] - 2xy = 0 ? (x-y)[sup:1ghphnax]2[/sup:1ghphnax] = 0 ? x - y = 0 ? x = y

Then

x[sup:1ghphnax]2[/sup:1ghphnax] + y[sup:1ghphnax]2[/sup:1ghphnax] = 1 ? x[sup:1ghphnax]2[/sup:1ghphnax] + x[sup:1ghphnax]2[/sup:1ghphnax] = 1 ? 2x[sup:1ghphnax]2[/sup:1ghphnax] = 1 ? x[sup:1ghphnax]2[/sup:1ghphnax] = 1/2 ? x = ± 1/?2

So the solutions are (1/?2 , 1/?2) and (-1/?2 , -1/?2)

thanks. that kind of helped me understand how x = y.