System of Equations Involving a Quadratic: Have Answer <> Understand

In summary, the system of equations can be solved using the substitution method, where we substitute x + y for v in the second equation. This leads to the quadratic equation 2x^2-2vx=0, which can be factored as 2x(x-v)=0. The possible solutions for x and y are x=0 and y=v.
  • #1
kwixson
6
0

Homework Statement



Solve this system of equations for x and y.

v=x+y
v^2=x^2+y^2

Homework Equations



The quadratic formula:

x = (-b +/- sqrt(b^2-4*a*c))/(2*a)

The Attempt at a Solution



A TA gave the following advice:

"Make y the subject of the first equation.
Find y2 in terms of v and x using this equation.
Substitute y2 in the second equation.
You now have a quadratic equation in x and there will be two solutions"​

I know the answers are x=0 and y=v but I "can't get there from here." My most recent attempt I got as far as 2x^2-2vx=0.

In this case, this homework is already solved so if someone could walk me through it I would be grateful.
 
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  • #2
kwixson said:

Homework Statement



Solve this system of equations for x and y.

v=x+y
v^2=x^2+y^2

Homework Equations



The quadratic formula:

x = (-b +/- sqrt(b^2-4*a*c))/(2*a)

The Attempt at a Solution



A TA gave the following advice:

"Make y the subject of the first equation.
Find y2 in terms of v and x using this equation.
Substitute y2 in the second equation.
You now have a quadratic equation in x and there will be two solutions"​

I know the answers are x=0 and y=v but I "can't get there from here." My most recent attempt I got as far as 2x^2-2vx=0.

In this case, this homework is already solved so if someone could walk me through it I would be grateful.

I would do something different from what your TA suggested. The first equation is v = x + y. I would substitute substitute x + y for v in the second equation, to get (x + y)2 = x2 + y2.

Expand the left side and simplify. What do you get?
 
  • #3
kwixson said:
I know the answers are x=0 and y=v but I "can't get there from here." My most recent attempt I got as far as 2x^2-2vx=0.

So is your question just how to solve this quadratic equation really? Because you can easily factor it
2x(x-v) = 0

So what are the possibly solutions?
 

Related to System of Equations Involving a Quadratic: Have Answer <> Understand

1. What is a system of equations involving a quadratic?

A system of equations involving a quadratic is a set of two or more equations that contain a quadratic term (x^2) and may also contain linear terms (x) and constant terms. These equations are solved simultaneously to find the values of x that satisfy all of the equations.

2. How do I solve a system of equations involving a quadratic?

To solve a system of equations involving a quadratic, you can use the substitution method, elimination method, or graphing method. First, rearrange the equations so that the terms are in the same order. Then, use one of the methods to solve for the value(s) of x that make the equations true.

3. Can a system of equations involving a quadratic have multiple solutions?

Yes, a system of equations involving a quadratic can have multiple solutions. This means that there can be more than one value of x that satisfies all of the equations in the system. These solutions can be real or complex numbers.

4. What does it mean to have an answer but not understand it in a system of equations involving a quadratic?

Having an answer but not understanding it in a system of equations involving a quadratic means that you have found a solution to the equations, but you do not fully understand the steps or reasoning behind the solution. This could be due to using a calculator or not fully comprehending the concepts behind solving quadratic equations.

5. How can I check if my solution is correct in a system of equations involving a quadratic?

To check if your solution is correct in a system of equations involving a quadratic, you can substitute the values of x into each equation and see if they make the equation true. If the values satisfy all of the equations, then your solution is correct. You can also check your solution by graphing the equations and seeing if the points of intersection match your solution.

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