Problem 5 - Precalculus: Mathematics for Calculus

Question

Solve the equation \(\sqrt{2 x}+x=0\) by doing the following steps. (a) Isolate the radical: _______ (b) Square both sides: ______ (c) The solutions of the resulting quadratic equation are ______ (d) The solution(s) that satisfy the original equation are _______

Step-by-Step Solution

Answer

The solution is \(x = 0\).

Step 1: Isolate the Radical

Let's begin by isolating the radical expression in the given equation. The original equation is \(\sqrt{2x} + x = 0\). To isolate \(\sqrt{2x}\), move \(x\) to the other side of the equation: \[\sqrt{2x} = -x\]

Step 2: Square Both Sides

Now, we will square both sides of the equation to eliminate the square root. Start with:\[ \left(\sqrt{2x}\right)^2 = (-x)^2 \]Simplifying both sides, the equation becomes:\[ 2x = x^2 \]

Step 3: Form a Quadratic Equation

The equation \(2x = x^2\) can be rearranged to form a standard quadratic equation:\[ x^2 - 2x = 0 \]This can be factored to:\[ x(x - 2) = 0 \]

Step 4: Solve the Quadratic Equation

Set each factor equal to zero to find the solutions:1. \(x = 0\)2. \(x - 2 = 0\) which gives \(x = 2\)

Step 5: Verify Solutions Against Original Equation

We need to substitute the solutions \(x = 0\) and \(x = 2\) back into the original equation \(\sqrt{2x} + x = 0\) to verify which are valid.- For \(x = 0\): \(\sqrt{2 \times 0} + 0 = 0\) is true.- For \(x = 2\): \(\sqrt{2 \times 2} + 2 = \sqrt{4} + 2 = 2 + 2 = 4 eq 0\) is false.Therefore, only \(x = 0\) is a valid solution.

Key Concepts

Radical Equations

Radical equations are equations where the unknown variable is inside a radical, usually a square root. Solving these equations typically involves isolating the radical to one side of the equation. This often makes it easier to solve because you can eliminate the radical by squaring both sides. This is exactly what we do with the equation \(\sqrt{2x} + x = 0\). By isolating \(\sqrt{2x}\), we transform the equation into a form that is easier to handle, such as \(\sqrt{2x} = -x\). It's important to be cautious during this process since squaring both sides might introduce extraneous solutions, which you'll need to check afterward.

Quadratic Equations

Quadratic equations follow the standard form \(ax^2 + bx + c = 0\). In our exercise, after squaring both sides, we ended up with \(2x = x^2\). This can be rewritten as \(x^2 - 2x = 0\), a quadratic equation. Quadratics can have up to two solutions, and understanding how to manipulate them is crucial for solving them effectively. The solutions to a quadratic equation are found, often using factoring, completing the square, or the quadratic formula. Here, factoring is the best approach. Quadratics often arise when dealing with squared relationships like we see here when the equation involves a radical.

Factoring Quadratics

Factoring quadratics is a method used to solve quadratic equations by rewriting the quadratic expression as a product of its factors. The equation \(x^2 - 2x = 0\) can be factored by taking out the common factor \(x\), resulting in \(x(x - 2) = 0\). This method is efficient, especially for simple quadratic equations where \(a = 1\) and allows us to quickly find the roots. Once factored, we apply the zero-product property that states if \(ab = 0\), then either \(a=0\) or \(b=0\). This lets us set each factor to zero, giving us the equations \(x = 0\) and \(x - 2 = 0\), leading to the solutions \(x = 0\) and \(x = 2\).

Solution Verification

Solution verification is the essential step of checking each solution to see if it satisfies the original equation. While solving \(\sqrt{2x} + x = 0\), we find the potential solutions \(x = 0\) and \(x = 2\). To verify these, substitute each back into the original equation. For \(x = 0\), the equation holds true: \(\sqrt{0} + 0 = 0\). However, for \(x = 2\), substituting it back yields \(\sqrt{4} + 2 = 4 eq 0\). This discrepancy indicates that \(x = 2\) is an extraneous solution created during the squaring process. Only \(x = 0\) is the true solution. This verification step is crucial because it confirms the validity of the solutions and ensures the solution is correct.