Problem 96 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{5 x+2}-\sqrt{x+10}=0 $$

Step-by-Step Solution

Answer

The solution is \(x = 2\).

Step 1: Isolate one square root

Start by moving one of the square root expressions to the other side of the equation: \( \sqrt{5x + 2} = \sqrt{x + 10} \). This simplifies our work moving forward.

Step 2: Square Both Sides to Eliminate the Square Roots

Square both sides of the equation to eliminate the square roots: \((\sqrt{5x + 2})^2 = (\sqrt{x + 10})^2\). This simplifies to \(5x + 2 = x + 10\).

Step 3: Solve the Resulting Linear Equation

Now solve the linear equation obtained: \(5x + 2 = x + 10\). Subtract \(x\) from both sides to get \(4x + 2 = 10\). Then, subtract 2 from both sides: \(4x = 8\), and finally divide by 4: \(x = 2\).

Step 4: Check for Extraneous Solutions

Substitute \(x = 2\) back into the original equation: \(\sqrt{5(2)+2} - \sqrt{2+10}\). This becomes \(\sqrt{12} - \sqrt{12} = 0\), which is true. Since our solution satisfies the original equation, it is not extraneous.

Key Concepts

Isolating Square Roots

When solving equations that involve square roots, our first step is to isolate one of the square root terms. This makes the equation easier to handle. In the given exercise, the original equation is \( \sqrt{5x + 2} - \sqrt{x + 10} = 0 \). To isolate one of the square roots, move one expression to the other side of the equation. Here, you move \( \sqrt{x + 10} \) over to give you \( \sqrt{5x + 2} = \sqrt{x + 10} \).

• Isolating a square root helps in neutralizing its impact, leaving us with another expression that can be further simplified.
• This step is about re-arranging terms until each side of the equation contains one square root or, best case, none.

By isolating the square roots, you can see clearly how the balance changes, preparing the equation for the next step: squaring.

Squaring Both Sides

Once the square roots have been isolated, the next step is to eliminate them. This is done by squaring both sides of the equation. Squaring is useful because the square of a square root returns the original term under the root. For our problem, you have after isolation: \( \sqrt{5x + 2} = \sqrt{x + 10} \). By squaring both sides, \[(\sqrt{5x + 2})^2 = (\sqrt{x + 10})^2\] results in: \[5x + 2 = x + 10\].

• Squaring removes the square root, which is crucial because it allows us to solve the resulting expression as a simple linear equation.
• Be cautious: squaring can introduce extraneous solutions, so we must be sure to check each solution at the end.

With the square roots cleared and the equation simplified, we're ready to tackle solving the linear equation.

Checking Solutions

No matter how spot-on your arithmetic seems, accuracy is vital, especially after squaring both sides which can produce extraneous solutions. An extraneous solution is a number that emerges from solving the equation but doesn’t satisfy the original equation. With our equation, after solving, we found \( x = 2 \). We check our work by putting \( x=2 \) back into the original equation: \( \sqrt{5(2) + 2} - \sqrt{2 + 10} = \sqrt{12} - \sqrt{12} \) This simplifies to \( 0 \). Since the left-hand side equals the right-hand side, our solution is verified and not extraneous.

• Always substitute your solution back into the original equation to ensure it holds true.
• This step might seem like extra work, but it prevents incorrect answers from being accepted.

Checking solutions is an essential safeguard in mathematics, especially when working with equations involving square roots.