Problem 15 - Algebra 2

Question

Find the exact solution(s) of each system of equations. $$ \begin{array}{l}{y+x^{2}=3} \\ {x^{2}+4 y^{2}=36}\end{array} $$

Step-by-Step Solution

Answer

The solutions are \((0, 3)\), \(\left(\frac{\sqrt{23}}{2}, -\frac{11}{4}\right)\), and \(\left(-\frac{\sqrt{23}}{2}, -\frac{11}{4}\right)\)."

Step 1: Express y in terms of x from Equation 1

The first equation is \( y + x^2 = 3 \). To express \( y \) in terms of \( x \), subtract \( x^2 \) from both sides to get: \( y = 3 - x^2 \).

Step 2: Substitute y in Equation 2

Now that we have \( y = 3 - x^2 \), substitute this expression into the second equation \( x^2 + 4y^2 = 36 \): \[ x^2 + 4(3 - x^2)^2 = 36 \]

Step 3: Simplify and Solve for x

First expand \( (3 - x^2)^2 \), which gives \( 9 - 6x^2 + x^4 \). Substitute into the equation:\[ x^2 + 4(9 - 6x^2 + x^4) = 36 \]Simplifying yields:\[ x^2 + 36 - 24x^2 + 4x^4 = 36 \]Combine like terms:\[ 4x^4 - 23x^2 = 0 \]Factor out \( x^2 \):\[ x^2(4x^2 - 23) = 0 \]Thus, either \( x^2 = 0 \) or \( 4x^2 = 23 \).

Step 4: Solve for x and y

If \( x^2 = 0 \), then \( x = 0 \). Substituting \( x = 0 \) in \( y = 3 - x^2 \) gives \( y = 3 \), so one solution is \( (0, 3) \).For \( 4x^2 = 23 \), solve for \( x \):\[ x^2 = \frac{23}{4} \]\[ x = \pm\sqrt{\frac{23}{4}} = \pm\frac{\sqrt{23}}{2} \].Substituting \( x = \frac{\sqrt{23}}{2} \) into \( y = 3 - x^2 \) gives \( y = 3 - \frac{23}{4} = \frac{-11}{4} \) and similarly for \( x = -\frac{\sqrt{23}}{2} \).Thus, the other solutions are \( \left(\frac{\sqrt{23}}{2}, -\frac{11}{4}\right) \) and \( \left(-\frac{\sqrt{23}}{2}, -\frac{11}{4}\right) \).

Key Concepts

Polynomial Equations

A polynomial equation is an equation that involves variables raised to whole number powers. These are equations like quadratic or cubic ones, but even higher degrees are possible. In the exercise at hand, both given equations are examples of polynomial equations.

• The first equation, \( y + x^2 = 3 \), is a simple example of how a polynomial can look with a lower degree term on one side of the equation.
• The second equation, \( x^2 + 4y^2 = 36 \), illustrates a system where both \( x \) and \( y \) variables are squared, making it a quadratic system. This results in a more complex interaction between the two variables.

Understanding polynomial equations is crucial since they lay the groundwork for solving more complex nonlinear systems of equations.

Substitution Method

The substitution method is a popular technique for solving systems of equations. In this approach, you first solve one of the equations for one variable and then substitute that expression into the other equation. This method simplifies the problem to finding the value of one variable, reducing computational complexity.
In the example provided, we first solve the equation \( y + x^2 = 3 \) for \( y \). By rearranging the equation, we express \( y \) as \( y = 3 - x^2 \). This expression is then substituted into the second equation, \( x^2 + 4y^2 = 36 \). This replaces \( y \) in the equation and allows us to solve for \( x \) only, reducing our two-variable system to a single-variable problem. Once \( x \) is determined, we use this value to find \( y \) using our first expression.

Factoring Equations

Factoring is a technique used to simplify mathematical expressions and solve equations. It involves writing the equation as a product of its factors, setting it equal to zero, and solving each factored term separately.
In this context, once we substitute and simplify the equation, we arrive at the equation \( 4x^4 - 23x^2 = 0 \). Factoring out \( x^2 \) from the equation shows us: \( x^2(4x^2 - 23) = 0 \).

• The solution \( x^2 = 0 \) gives us \( x = 0 \), which is straightforward.
• The other equation \( 4x^2 - 23 = 0 \) allows us to solve for \( x^2 = \frac{23}{4} \), which results in two solutions for \( x \), due to the nature of square roots, \( x = \pm \frac{\sqrt{23}}{2} \).

Factoring is essential as it breaks down complex polynomial expressions into manageable pieces that we can solve step by step.

Solution Verification

After solving the equations using substitution and factoring, it's crucial to verify the solutions to ensure they satisfy both original equations. Verification not only confirms the correctness of the derived solutions but also highlights any errors made in the calculations.
In our scenario, we found three potential solutions \((0, 3)\), \(\left(\frac{\sqrt{23}}{2}, -\frac{11}{4}\right)\), and \(\left(-\frac{\sqrt{23}}{2}, -\frac{11}{4}\right)\). To verify:

• Substitute \((0, 3)\) into both original equations to confirm they hold true.
• Do the same for \(\left(\frac{\sqrt{23}}{2}, -\frac{11}{4}\right)\) and check both equations' fulfillment.
• Lastly, verify the solution \(\left(-\frac{\sqrt{23}}{2}, -\frac{11}{4}\right)\).

Always make it a practice to verify solutions to catch errors early and build confidence in the reliability of your results.