Problem 37 - College Algebra

Question

Solve each system by the method of your choice. $$\begin{aligned} &x^{2}+(y-2)^{2}=4\\\ &x^{2}-2 y=0 \end{aligned}$$

Step-by-Step Solution

Answer

The solutions for the system of equations are (0, 0) and (2, 2), (-2, 2)

Step 1: Expressing 'x' in terms of 'y' from the second equation

The second equation is more straightforward, hence rearranging terms can be done to express 'x' in terms of 'y'. Rearranging \(x^{2}-2y=0\), \(x^{2}=2y\), which implies \(x = \sqrt{2y}\)

Step 2: Substituting x into the first equation

Substitute \(x = \sqrt{2y}\) in \(x^{2}+(y-2)^{2}=4\). This results into \( (\sqrt{2y})^{2}+(y-2)^{2}=4\) which simplifies to \(2y+(y-2)^{2}=4\)

Step 3: Solving the equation for 'y'

Solve the quadratic equation \(2y+(y-2)^{2}=4\). Simplifying, we get: \(y^{2}-2y+4=4\), which simplifies to \(y^{2}-2y=0\), and from there we find \(y=0\) or \(y=2\) by factorization method.

Step 4: Calculating the value of 'x' for each 'y' value

Now that we have the y values, we can substitute them into the equation \(x = \sqrt{2y}\) to find the corresponding x values. For \(y=0\), \(x = \sqrt{2*0} = 0\) For \(y=2\), \(x = \sqrt{2*2} = 2\) or \(x = -\sqrt{2*2} = -2\)

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental concept in algebra, representative of polynomial equations of degree two. These equations generally take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). Solving quadratic equations is crucial because they appear frequently in various real-world problems, such as calculating areas or dealing with physics problems. Two common methods for solving them are the quadratic formula and factoring. For the exercise at hand, the quadratic component arises when we solve for \(y\) in the expression \((y-2)^2\), resulting from expanding squares in the main equation. The equation \((y-2)^2 = 4\) nonlinearizes the system, requiring careful manipulation to solve.

Substitution Method

The substitution method is a strategic approach for solving systems of equations, especially when one equation is more straightforward. This method involves solving one of the equations for a single variable and then substituting this expression into the other equation.
For example, in the exercise, we start by using the second equation \(x^2 - 2y = 0\), which can be rearranged to express \(x^2\) in terms of \(y\): \(x^2 = 2y\).
Once the expression for \(x\) is isolated, you substitute \(\sqrt{2y}\) in place of \(x\) in the first equation. This reduces the system to a simpler form that can be more easily solved. Through substitution, a system that initially involves multiple variables becomes a single equation with one variable, simplifying the solving process significantly.

Solving for Variables

Solving for variables involves finding the specific numerical values that satisfy all equations in a system. In systems involving quadratic equations, like the one we are dealing with, this often requires overcoming non-linear expressions.
After applying the substitution method to reduce the number of variables, you then proceed to solve the quadratically transformed equation for \(y\). For instance, through expansion and simplification, our equation becomes \(y^2 - 2y = 0\).
To solve such a quadratic equation, one could use factoring, yielding solutions for \(y\) like \(y = 0\) and \(y = 2\). Once \(y\) values are known, we revert to the substituted expression to find corresponding \(x\) values. Hence, for \(y = 0\), \(x = 0\); and for \(y = 2\), both \(x = 2\) and \(x = -2\) are calculated, clearly illustrating the relationship between solving for one variable and finding solutions for others.