Problem 51 - Algebra 1: Concepts and Skills

Question

Solve the equation. Check for extraneous solutions. $$ x=\sqrt{2 x+3} $$

Step-by-Step Solution

Answer

The solution for the equation is \(x = 3\). After squaring both sides of the equation and rearranging it into a quadratic form, the quadratic equation is solved to give two potential solutions \(x = 3, -1\). However, when they are verified by substituting back into the original equation, \(x = -1\) proved to be an extraneous solution. So, the only valid solution for the original equation is \(x = 3\).

Step 1: Squaring both sides

Start by squaring both sides of the equation to get rid of the square root in the right side of the equation. This will give: \(x^2=(\sqrt{2x+3})^2\) , which simplifies to \(x^2=2x+3\).

Step 2: Form a quadratic equation

Now, bring all terms to one side to create a quadratic equation. This translates to: \(x^2-2x-3=0\).

Step 3: Solve the quadratic equation

The quadratic equation can be solved by factoring, using the quadratic formula or completing the square. In this case factoring is the easiest method. We need two numbers that multiply to -3 and add to -2 which are -3 and 1. Hence, the factored form of the equation will be \((x-3)(x+1)=0\). Hence, potential solutions are \(x=3, -1\).

Step 4: Check for extraneous solutions

After finding the potential solutions, we need to substitute each one of them back into the original equation to check if they are valid solutions or if they are extraneous solutions. If we substitute \(x=3\), we get \(\sqrt{2*3 + 3} = 3\) which simplifies to \(\sqrt{9}=3\), hence \(x=3\) is a valid solution. On the other hand, substituting \(x=-1\) into the original equation, we would get \(\sqrt{2*-1 +3} = -1\) which simplifies to \( \sqrt{1} = -1\), which is not true. Hence, \(x=-1\) is an extraneous solution.

Key Concepts

Quadratic Equations

Quadratic equations involve expressions that are set equal to zero with a variable to the second power, generally represented as \( ax^2 + bx + c = 0 \). In our exercise, after removing the square root from the equation \( x = \sqrt{2x + 3} \), we developed a quadratic equation by squaring both sides, leading to \( x^2 = 2x + 3 \). The next crucial step is aligning terms to reach the standard form of the quadratic equation, resulting in \( x^2 - 2x - 3 = 0 \). Solving quadratic equations like this can often be done using methods such as:

• Factoring - the method used in our solution.
• Quadratic formula - \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Completing the square - rewriting it into a perfect square trinomial.

The goal is to find values of \( x \) that satisfy the equation.

Extraneous Solutions

Extraneous solutions are false solutions that appear when manipulating equations, particularly when squaring both sides. Therefore, it's crucial to verify each possible solution by substituting it back into the original equation.Here's why extraneous solutions happen:

• When you square both sides of an equation, you lose track of the sign of the original expression on the right side. In this exercise, although both \( x = 3 \) and \( x = -1 \) solved the factored quadratic, substituting \( x = -1 \) back resulted in an incorrect equation: \( \sqrt{1} eq -1 \).

Always substitute potential solutions back into the original equation to confirm their validity. This avoids accepting solutions that don't work numerically or logically.

Factoring Methods

Factoring is a straightforward method to solve quadratic equations by expressing them as a product of two binomials. Consider the factored form of the equation \( x^2 - 2x - 3 = 0 \) as \((x - 3)(x + 1) = 0\).The key steps in factoring are:

• Find two numbers that multiply to the constant term \( c \), here \(-3\), and add to the linear coefficient \( b \), here \(-2\).
• The numbers \(-3\) and \(+1\) satisfy these conditions because \(-3 \times 1 = -3\) and \(-3 + 1 = -2\).
• Use these numbers to write the factored form: \((x - 3)(x + 1)\).

Lastly, set each factor to zero, \( x - 3 = 0 \) or \( x + 1 = 0 \), leading to the solutions \( x = 3 \) and \( x = -1 \). Factoring is efficient and effective, especially when the quadratic is simple to manipulate.