Problem 38 - Precalculus : Building Concepts and Connections

Question

Solve the radical equation to find all real solutions. Check your solutions. $$\sqrt{x^{2}+3}=\sqrt{28}$$

Step-by-Step Solution

Answer

The solution to the equation \( \sqrt{x^{2}+3}=\sqrt{28}\) is \(x = -5, x = 5\). Both solutions were checked and hold true for the original radical equation.

Step 1: Square both sides of the equation

To eliminate the square roots, square both sides of the equation. When you square a square root, the two operations cancel each other out. So you get \(x^{2}+3=28\)

Step 2: Solve for the variable x

First, move the constant on the left to the right side to isolate the variable term on one side, which gives you \(x^{2}=28-3\). Simplifying the right side gives \(x^{2}=25\). Now, taking the square root of 25 on the right side will give two values, one positive and one negative, because both -5 and 5 when squared, give 25. Therefore, \(x = \pm 5\).

Step 3: Check the solutions

Substitute both \(x = -5\) and \(x = 5\) into the original equation and see if both sides of the equation are equal. With \(x = -5\), you plug it in the equation and get \(\sqrt{(-5)^{2}+3} = \sqrt{28}\). Upon simplification, you get \(\sqrt{28} = \sqrt{28}\), which holds true. With \(x = 5\), the process is similar and also returns a true statement, so both solutions are valid.

Key Concepts

Square Roots Manipulation

Understanding how to manipulate square roots is crucial in solving equations involving radicals. A square root symbol, \( \sqrt{} \), essentially asks for a number that, when multiplied by itself, will yield the original number inside the root. An important property to remember is that \( \sqrt{a^2} = |a| \), which acknowledges the absolute value since both positive and negative numbers squared will give a positive result.

When faced with an equation like \( \sqrt{x^2 + 3} = \sqrt{28} \), the first instinct should be to eliminate the square roots to simplify the problem. This is done by squaring both sides, taking care to apply the exponent to the entire contents of the square root. After this step, you are often left with a standard algebraic equation to solve. However, one must be mindful during the step of squaring as it might introduce extraneous solutions. That's why checking the solutions later is an essential part of the process.

Isolate the Variable

Isolating the variable is a foundational step in solving almost any algebraic equation. The goal is to have the variable on one side of the equation and all other terms on the opposite side. This process often involves basic arithmetic operations: addition or subtraction to move terms across the equal sign and multiplication or division to consolidate or disperse coefficient values.

For instance, for the squared form of our equation \(x^2 + 3 = 28\), you would subtract 3 from both sides to get \(x^2 = 25\). Reaching this point sets you up perfectly to then perform the square root operation to find the value(s) of \(x\). It's vital to proceed with these steps logically and methodically, checking for any mistakes at each stage. An algebraic equation is a balance; what you do to one side, you do to the other to maintain equality.

Checking Solutions in Equations

After finding potential solutions, especially in radical equations, it is imperative to check each one to ensure that they actually satisfy the original equation. Some operations, like squaring both sides of an equation, can introduce solutions that don't hold true when plugged back in. This process of validation is known as 'checking your solutions.'

To check a solution, you substitute the solution back into the original equation and see if the equation 'balances' or makes sense. In our example, after determining that \( x = \pm 5 \), we replaced \( x \) with 5 and -5 in the original equation and confirmed that both solved the equation accurately. This final step is not just a formality; it actually confirms that the operations performed did not disturb the integrity of the original equation.