Problem 43 - Algebra 1

Question

Solve the equation. Check for extraneous solutions. $$\sqrt{110-x}=x$$

Step-by-Step Solution

Answer

The solution to the equation is \(x = 10\).

Step 1: Squaring both sides

To eliminate the square root, we should square both sides of the equation. By doing so, we get the equation: \((\sqrt{110-x})^2 = x^2\). Simplifying this produces the equation: \(110-x = x^2\).

Step 2: Rewrite as a quadratic equation

We can rewrite the equation as a quadratic equation in the standard form \(ax^2 + bx + c = 0\). Here, rearranging the equation gives: \(x^2 + x - 110 = 0\).

Step 3: Solve the quadratic equation

We can now solve for \(x\) using the quadratic formula: \(x = [-b \pm \sqrt{(b^2 - 4ac)}] / (2a)\). Substituting \(a=1\), \(b=1\) and \(c=-110\) into the formula, we find two potential solutions: \(x = -11\) and \(x = 10\).

Step 4: Check for extraneous solutions

We now substitute each of these potential solutions back into the original equation to verify if they are valid. When \(x = -11\), we obtain \(\sqrt{110 - (-11)} = -11\), which simplifies to \(\sqrt{121} = -11\). This is not true and thus \(x = -11\) is an extraneous solution. When \(x = 10\), we obtain \(\sqrt{110 - 10} = 10\), which simplifies to \(\sqrt{100} = 10\). This is true and thus \(x = 10\) is a valid solution.

Key Concepts

Extraneous Solutions

An important concept when solving equations, especially those involving square roots, is identifying extraneous solutions. Extraneous solutions can occur when we apply certain mathematical operations, such as squaring both sides of an equation.
These operations can introduce solutions that weren't originally present in the physical or original problem context.

• When you square both sides of an equation to remove a square root, you must check each derived solution in the original equation.
• Extraneous solutions do not satisfy the original equation.

In the exercise provided, squaring both sides creates a quadratic equation. Solving it can result in solutions that appear mathematically correct, but one must manually verify these solutions. This ensures that they satisfy the original problem, such as by substituting back into the original equation. By doing this, we avoid accepting an extraneous solution as valid.

Solving Equations

Solving equations often involves a series of strategic steps to simplify and isolate variables. Here’s a simple breakdown of the process:

• Identify the equation type (linear, quadratic, etc.). For the given exercise, it's a radical equation.
• Apply operations that simplify the equation. Here, squaring was used to eliminate the square root.
• Rearrange the terms to bring the equation into a standard form suitable for further solving, like bringing it to the quadratic form.

Another crucial part of solving equations is keeping track of assumptions and restrictions. For instance, after squaring both sides, checking the validity of potential solutions against the original function ensures that the logical structure of the problem is maintained. Ultimately, solving equations is about balancing both sides until you isolate the variables and determine their exact values.

Quadratic Formula

The quadratic formula is a reliable tool for solving quadratic equations, which take the form \[ ax^2 + bx + c = 0. \] It is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]
This powerful method finds the roots of any quadratic equation, where

• \( a \), \( b \), and \( c \) are coefficients from the equation.
• The expression under the square root, \( b^2 - 4ac \), is called the discriminant. It tells us about the nature of the roots.
• If the discriminant is positive, the quadratic has two distinct real roots.

For the exercise at hand, using the quadratic formula provided two values:\( x = 10 \) and \( x = -11 \).
However, only \( x = 10 \) was valid upon checking back with the original equation. The quadratic formula is especially useful because it can solve any quadratic, even when other factoring methods fail.