Problem 61 - College Algebra with Modeling and Visualization

Question

If possible, solve the nonlinear system of equations. $$ \begin{array}{r} \sqrt{x}-2 y=0 \\ x-y=-2 \end{array} $$

Step-by-Step Solution

Answer

The solutions are \( \left( \frac{1+\sqrt{33}}{8} - 2, \frac{1+\sqrt{33}}{8} \right) \) and \( \left( \frac{1-\sqrt{33}}{8} - 2, \frac{1-\sqrt{33}}{8} \right) \).

Step 1: Solve the second equation for x

Let's first solve the equation \( x - y = -2 \) for \( x \). By adding \( y \) to both sides, we get \( x = y - 2 \). Now we have expressed \( x \) in terms of \( y \).

Step 2: Substitute x in the first equation

Substitute \( x = y - 2 \) from Step 1 into the first equation \( \sqrt{x} - 2y = 0 \). This gives us: \[ \sqrt{y-2} - 2y = 0 \].

Step 3: Isolate the square root

Add \( 2y \) to both sides to isolate the square root: \[ \sqrt{y-2} = 2y \].

Step 4: Square both sides to eliminate the square root

Square both sides to solve for \( y \): \[ (\sqrt{y-2})^2 = (2y)^2 \] which simplifies to \[ y-2 = 4y^2 \].

Step 5: Rearrange into a standard quadratic form

Rearrange the equation from Step 4 into a standard quadratic form: \[ 4y^2 - y - 2 = 0 \].

Step 6: Solve the quadratic equation for y

Solve \( 4y^2 - y - 2 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a=4 \), \( b=-1 \), \( c=-2 \). Calculate the discriminant: \( b^2 - 4ac = 1 - 4(4)(-2) = 1 + 32 = 33 \). Thus, \[ y = \frac{1 \pm \sqrt{33}}{8} \].

Step 7: Find corresponding x values

For each solution for \( y \), find \( x \) using \( x = y - 2 \). For \( y = \frac{1 + \sqrt{33}}{8} \), \( x = \frac{1 + \sqrt{33}}{8} - 2 \). Similarly, calculate \( x \) for \( y = \frac{1 - \sqrt{33}}{8} \) as \( x = \frac{1 - \sqrt{33}}{8} - 2 \).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation. They come in the standard form \( ax^2 + bx + c = 0 \). Here, "a", "b", and "c" are constants, and "x" is the variable. Importantly, "a" should not be zero, because that's what makes the equation quadratic. The general shape of the graph for a quadratic equation is a parabola, which can open upwards or downwards depending on the sign of the coefficient "a".

To solve quadratic equations, you can use the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

The quadratic formula is derived from completing the square method and is a straightforward way to find the roots of a quadratic equation given the coefficients. The discriminant, \( b^2 - 4ac \), inside the square root tells us the nature of the roots:

• If it's positive, there are two distinct real roots.
• If it's zero, there's exactly one real root.
• If negative, the roots are complex.

Systems of Equations

A system of equations involves two or more equations with multiple variables. The goal is to find a set of values for the variables that satisfies all equations in the system simultaneously. One common approach is the substitution method, particularly useful when one equation is easily solvable for one variable.

In our example, the system includes:

• \( \sqrt{x} - 2y = 0 \)
• \( x - y = -2 \)

The second equation was simplified to express \( x \) in terms of \( y \), making it straightforward to substitute into the first equation. This approach efficiently reduces the problem to a single variable equation, making it simpler to solve. Another method is elimination, but it requires some modifications when dealing with nonlinear equations.

Square Root Property

The square root property is a useful algebraic tool when solving equations with square roots. It states that if \( a = b^2 \), then \( b = \pm\sqrt{a} \). This method is valuable for eliminating square roots from equations, transforming them into more solveable forms.

In the given system, we isolated the square root in the equation \( \sqrt{y-2} = 2y \). By squaring both sides, we eliminated the square root and reduced it to a polynomial equation, specifically a quadratic equation: \( y - 2 = 4y^2 \).

Squaring is a powerful but cautious step. You should always be aware that it might introduce extraneous solutions. These are solutions that fit the squared version but not the original equation. When you finally solve the quadratic equation, always verify solutions by plugging them back into the original problem to ensure their validity.