Problem 28 - Algebra and Trigonometry

Question

\(19-32\) . Find all solutions of the system of equations. $$ \left\\{\begin{aligned} x^{2}+2 y^{2} &=2 \\ 2 x^{2}-3 y &=15 \end{aligned}\right. $$

Step-by-Step Solution

Answer

Solutions are \((\sqrt{9.75}, 1.5), (-\sqrt{9.75}, 1.5), (\sqrt{4.125}, -2.25), (-\sqrt{4.125}, -2.25)\).

Step 1: Simplify Second Equation for y

Starting with the second equation, express it in terms of one variable. You have:\[2x^2 - 3y = 15\].Rearrange this to solve for \(y\):\[3y = 2x^2 - 15\],which gives us:\[y = \frac{2x^2 - 15}{3}\].

Step 2: Substitute y into First Equation

Substitute \(y = \frac{2x^2 - 15}{3}\) into the first equation \(x^2 + 2y^2 = 2\). This leads to:\[x^2 + 2\left(\frac{2x^2 - 15}{3}\right)^2 = 2\].

Step 3: Simplify and Solve for x

Simplify the equation:\[x^2 + 2\left(\frac{4x^4 - 60x^2 + 225}{9}\right) = 2\],which becomes:\[x^2 + \frac{8x^4 - 120x^2 + 450}{9} = 2\].Eliminate the fraction by multiplying everything by 9:\[9x^2 + 8x^4 - 120x^2 + 450 = 18\].Further simplify to:\[8x^4 - 111x^2 + 432 = 0\].

Step 4: Solve the Polynomial Equation

Let \(z = x^2\), transforming the equation to:\[8z^2 - 111z + 432 = 0\].This is a quadratic equation in terms of \(z\). Use the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 8\), \(b = -111\), \(c = 432\):\[z = \frac{111 \pm \sqrt{111^2 - 4 \cdot 8 \cdot 432}}{16}\].Calculate under the square root:\[111^2 - 4 \cdot 8 \cdot 432 = 2025\],thus:\[z = \frac{111 \pm 45}{16}\].This results in:\[z_1 = \frac{111 + 45}{16} = 9.75\],\[z_2 = \frac{111 - 45}{16} = 4.125\].Since \(z = x^2\), find \(x\) values:\(x = \pm\sqrt{9.75}\) and \(x = \pm\sqrt{4.125}\).

Step 5: Substitute x values back into y

For each \(x\) found, return to the modified equation for \(y\):For \(x = \pm\sqrt{9.75}\): \(y = \frac{2(9.75) - 15}{3} = 1.5\).For \(x = \pm\sqrt{4.125}\): \(y = \frac{2(4.125) - 15}{3} = -2.25\).

Step 6: Find All Pairs

Combine all possible \((x, y)\) pairs. These are:1. \((\sqrt{9.75}, 1.5)\),2. \((-\sqrt{9.75}, 1.5)\),3. \((\sqrt{4.125}, -2.25)\),4. \((-\sqrt{4.125}, -2.25)\).

Key Concepts

Quadratic Equations

Quadratic equations are a special form of polynomial equations of degree 2. They follow the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). In solving systems of equations, quadratic equations frequently appear. They can be expressed in various forms but are always recognizable by the squared term. For instance, in our given exercise, both equations involve quadratic terms, with the first equation being \( x^2 + 2y^2 = 2 \). This equation tells us that the relationship between \( x \) and \( y \) includes squared terms, which are pivotal in defining the curve on the graph. To solve such quadratic equations, it often helps to isolate one of the squared terms, allowing for substitution method to solve the system. Identifying and understanding these equations is crucial, as they represent the core structure that must be balanced among the variables involved.

Substitution Method

The substitution method is a powerful technique for solving systems of equations, especially when dealing with non-linear equations like quadratics. It involves expressing one variable in terms of the other and then substituting this expression into another equation. In the exercise, we used the substitution method by first rearranging the second equation \( 2x^2 - 3y = 15 \) to solve for \( y \):

• Rearrange: \( 3y = 2x^2 - 15 \)
• Solve: \( y = \frac{2x^2 - 15}{3} \)

This expression for \( y \) is then substituted into the first equation \( x^2 + 2y^2 = 2 \). This replacement reduces the system to a single equation with only one unknown, \( x \). By performing these steps, we make it simpler to tackle the equation using algebra to expose the underlying relationships between the variables. The key advantage of substitution is its simplicity in handling complex systems, turning them into more manageable forms.

Solving Polynomial Equations

Polynomial equations can be intimidating due to their complexity, particularly when they involve higher degrees. However, they can be solved systematically. These equations are characterized by terms with variables raised to power and various constants. In the provided exercise, after using substitution, we arrive at the polynomial equation \( 8x^4 - 111x^2 + 432 = 0 \). To solve this, it was helpful to make a substitution again, setting \( z = x^2 \). This converts the equation into a quadratic form \( 8z^2 - 111z + 432 = 0 \), which can then be solved using the quadratic formula:

• Formula: \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• In this case: \( a = 8 \), \( b = -111 \), \( c = 432 \)

The solution to this gives values for \( z \), which are then back-transformed to find \( x \). This approach splits the complexity into manageable steps, making it easier to deal with variables raised to any power. Understanding how to handle polynomial equations is essential in solving any system of equations, especially those involving non-linear relationships.