How Do I Solve This Multivariable Algebraic Equation?

[dlacombe13]

Okay, so this equation is coming from a physics problem, which I essentially already solved (the physics part). The only thing I am stuck on is the algebra (I think) portion of it, so the details of the problem itself do not matter. I know the answer already as well, I just can't get to it myself. The equation is:
5v₁² - 3v_o²-2v₁v_o=0
For simplicity you can use:
5x²-3y²-2xy=0
I am solving for v₁ which should equal (-3v_o)/5
If someone could just point me in some direction, that would be great. I have no idea where to start, since I have very little experience with solving for multi-variables without 2 equations.

 

[cnh1995]

5x2-3y2-2xy=0

This is a quadratic equation for two variables. Look for the two numbers whose sum is equal to the coefficient of the xy term and whose product is equal to the product of coefficients of the square terms.

 

[SammyS]

Okay, so this equation is coming from a physics problem, which I essentially already solved (the physics part). The only thing I am stuck on is the algebra (I think) portion of it, so the details of the problem itself do not matter. I know the answer already as well, I just can't get to it myself. The equation is:
5v₁² - 3v_o²-2v₁v_o=0
For simplicity you can use:
5x²-3y²-2xy=0
I am solving for v₁ which should equal (-3v_o)/5
If someone could just point me in some direction, that would be great. I have no idea where to start, since I have very little experience with solving for multi-variables without 2 equations.

Item #1: Please use the template which is provided for you when starting a new thread.

Homework Statement

Homework Equations

The Attempt at a Solution

​

One technique for attacking your problem:

Divide both sides of the equation by v₀², or equivalently by y² .

You then have a quadratic equation in (v₁/v₀) or (x/y) .

If it's still not clear, replace either of those with some other variable, such as "u".

 

[Ray Vickson]

Okay, so this equation is coming from a physics problem, which I essentially already solved (the physics part). The only thing I am stuck on is the algebra (I think) portion of it, so the details of the problem itself do not matter. I know the answer already as well, I just can't get to it myself. The equation is:
5v₁² - 3v_o²-2v₁v_o=0
For simplicity you can use:
5x²-3y²-2xy=0
I am solving for v₁ which should equal (-3v_o)/5
If someone could just point me in some direction, that would be great. I have no idea where to start, since I have very little experience with solving for multi-variables without 2 equations.

You have a single quadratic equation for the single variable $v_1$, because your $v_0$ is just some input parameter, not a variable (at least, not in the way you described it). So, just use the familiar quadratic solution formula, and simplify it down as much as possible. You ought to find a second root besides the one $v_1 = -3v_0/5$ that you have given.

 

[dlacombe13]

Thank you all for your help. I searched for a while and finally found a method to factoring these types of equations. It has been so long since I've had to solve an equation like this, but it all makes sense now. I used a method called the box method to find the factors which ended up being (5x+3y)(x-y), which gave me the answers x=y, and x=(-3y)/5. I also tried dividing all terms by y² which gave me the equation in terms of x/y. I then subbed in 'u' for x/y which gave me:
5u² - 2u -3 = 0, which I also factored and then re-subbed x/y for 'u' to get the same results.

One final question to further my learning: How exactly would I use the quadratic formula to solve this? I tried plugging in the 'u' equation and got -3/5, but how would I sub back in (x/y) to get it in terms of y?

 

[dlacombe13]

Thank you all for your help. I searched for a while and finally found a method to factoring these types of equations. It has been so long since I've had to solve an equation like this, but it all makes sense now. I used a method called the box method to find the factors which ended up being (5x+3y)(x-y), which gave me the answers x=y, and x=(-3y)/5. I also tried dividing all terms by y² which gave me the equation in terms of x/y. I then subbed in 'u' for x/y which gave me:
5u² - 2u -3 = 0, which I also factored and then re-subbed x/y for 'u' to get the same results.

One final question to further my learning: How exactly would I use the quadratic formula to solve this? I tried plugging in the 'u' equation and got -3/5, but how would I sub back in (x/y) to get it in terms of y?

 

[PeroK]

Thank you all for your help. I searched for a while and finally found a method to factoring these types of equations. It has been so long since I've had to solve an equation like this, but it all makes sense now. I used a method called the box method to find the factors which ended up being (5x+3y)(x-y), which gave me the answers x=y, and x=(-3y)/5. I also tried dividing all terms by y² which gave me the equation in terms of x/y. I then subbed in 'u' for x/y which gave me:
5u² - 2u -3 = 0, which I also factored and then re-subbed x/y for 'u' to get the same results.

One final question to further my learning: How exactly would I use the quadratic formula to solve this? I tried plugging in the 'u' equation and got -3/5, but how would I sub back in (x/y) to get it in terms of y?

Let's compare two equations:

$x^2 + 2bx - 1 = 0$

And

$x^2 + 2yx - 1 = 0$

So, what precisely is the difference? The first is a quadratic in $x$ with an arbitrary parameter $b$. The second is an equation in two "variables" $x$ and $y$.

But, algebraically, you can solve both equations by the same method. Just complete the square (or apply the quadratic formula) to get:

$x = -b \pm \sqrt{b^2 + 1}$

And

$x = -y \pm \sqrt{y^2 + 1}$

So, what's the difference? Both are quadratics in $x$. Isn't it just different notation to use $b$ or $y$?

 

[dlacombe13]

Oh okay. So if I am following correctly:
$5x^2 - 2xy - 3y^2$

In terms of y:
$a = 5$
$b = -2y$
$c = -3y^2$

$x = \frac{2y\pm \sqrt{4y^2 - 4(5)(-3y^2)}}{10}$
$\Rightarrow x = \frac{2y\pm \sqrt{64y^2}}{10}$
$\Rightarrow x = \frac{2y \pm 8y}{10}$
$\Rightarrow x = [y, \frac{-3y}{5}$]

 

[PeroK]

Oh okay. So if I am following correctly:
$5x^2 - 2xy - 3y^2$

In terms of y:
$a = 5$
$b = -2y$
$c = -3y^2$

$x = \frac{2y\pm \sqrt{4y^2 - 4(5)(-3y^2)}}{10}$
$\Rightarrow x = \frac{2y\pm \sqrt{64y^2}}{10}$
$\Rightarrow x = \frac{2y \pm 8y}{10}$
$\Rightarrow x = y$
$\Rightarrow x = \frac{-3y}{5}$

Yes, algebra works with any letters!

 

[dlacombe13]

Thank you very much for your help! I know that it works for any letters, I just wasn't sure if it was legal to include $y$ as part of the coefficients.