Problem 26 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{aligned} &x=\sqrt{y}\\\ &x^{2}-y^{2}=4 \end{aligned}$$

Step-by-Step Solution

Answer

The solution to the system of equations is \((x, y) = (\sqrt{2}, 2)\).

Step 1: Substitute x with √y in the second equation

Replace x with √y in the second equation: \((\sqrt{y})^2 - y^2 = 4\)

Step 2: Simplify the resulting equation

Simplify the equation to get a quadratic equation in terms of y: \(y - y^2 = 4\)

Step 3: Rewriting the quadratic equation in standard form

Rearrange the equation to have the standard form of a quadratic equation: \(y^2 - y - 4 = 0\)

Step 4: Solve the quadratic equation for y

Factor the equation or use the quadratic formula to find the solutions for y: \((y - 2)(y + 2) = 0\) So, the possible values for y are: \(y = 2\) or \(y = -2\)

Step 5: Find the corresponding x values for each y value

For each y value, use the first equation to find the corresponding x value: For \(y = 2\): \(x = \sqrt{2}\) For \(y = -2\): x cannot be a real number since √(-2) is an imaginary number.

Step 6: Write the solution

Now that we have found the values of x and y that satisfy both equations, write the solution as an ordered pair: \((x, y) = (\sqrt{2}, 2)\) The solution to the system of equations is \((x, y) = (\sqrt{2}, 2)\).

Key Concepts

Quadratic Equations

Quadratic equations are essential to solving systems that include second-degree polynomials. These equations have the standard form: \[ ax^2 + bx + c = 0 \]where \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). In the context of solving our exercise, the equation evolved into a quadratic as we manipulated the terms to isolate powers of \(y\). After substituting \(x = \sqrt{y}\) into the system and simplifying, we obtained \[ y^2 - y - 4 = 0. \]This is a classic quadratic equation, which we can solve using various methods such as factoring, completing the square, or using the quadratic formula. Understanding quadratic equations is crucial as they frequently appear in systems where parabolas or curves are involved. Practice recognizing standard forms and applying the correct solution strategies is key in handling quadratic equations successfully.

Substitution Method

The substitution method is a popular technique for solving systems of equations. Its main goal is to express one variable in terms of the other, reducing the complexity of the system. In this method, once one variable is isolated, it's substituted or replaced into the other equation. This reduces the system from two equations to a single equation in one variable.

In our exercise, we had the equations:

• \(x = \sqrt{y}\)
• \(x^2 - y^2 = 4\)

We first expressed \(x\) as \(\sqrt{y}\) from the first equation. Then, substituted this into the second equation to simplify it:\[(\sqrt{y})^2 - y^2 = 4 \]This substitution allowed us to work with a single-variable quadratic equation, making it easier to solve. Mastering the substitution method is fundamental because it simplifies complex systems into more manageable single-variable equations.

Real Numbers

Real numbers include all the numbers that can be found on the number line. This set comprises both rational and irrational numbers. Rational numbers are those that can be expressed as a fraction, like 2 or -3/4. Irrational numbers cannot be expressed exactly as a simple fraction, like \(\sqrt{2}\) or \(\pi\).

In our system, the goal was to identify solutions that are real numbers. Since \(x = \sqrt{y}\), to keep \(x\) real, \(y\) must be non-negative. That is why we have real solutions for \(y = 2\) where \(x = \sqrt{2}\), ensuring both \(x\) and \(y\) are real numbers. Real number solutions are vital for finding solutions on the coordinate plane for real-world applications.

Imaginary Numbers

Imaginary numbers arise when we take the square root of a negative number. This is because real numbers cannot capture the essence of a square root of a negative, as they don't exist on the real number line. Instead, we define them using the imaginary unit \(i\), where \(i = \sqrt{-1}\).

In the problem, when we computed \(x = \sqrt{-2}\) for \(y = -2\), \(x\) was not a real number. In these situations, the solution becomes complex, involving the imaginary unit. Imaginary numbers are crucial to expanding the concept of a "number" beyond the real number line and are widely used in fields like engineering and physics for various applications, including circuit analysis and signal processing. Recognizing when a number is imaginary allows for comprehensive problem-solving, particularly in mathematical contexts involving complex equations.