Problem 23 - Algebra 2

Question

Solve. Check for extraneous solutions. \((7 x+6)^{\frac{1}{2}}=(9+4 x)^{\frac{1}{2}}\)

Step-by-Step Solution

Answer

The solution to the equation is \( x = 1 \).

Step 1: Square Both Sides

To eliminate the roots, square both sides of the equation: \( ((7x + 6)^{\frac{1}{2}})^2 = ((9 + 4x)^{\frac{1}{2}})^2 \) which simplifies to \( 7x + 6 = 9 + 4x \)

Step 2: Solve for x

To isolate the variable, subtract \( 4x \) from each side to get \( 3x + 6 = 9 \) and then subtract 6 from each side to get \( 3x = 3 \) Finally, divide by 3 to get \( x = 1 \)

Step 3: Check for Extraneous Solutions

Substitute \( x = 1 \) into the original equation. Substituting \( x = 1 \) into \( (7x + 6)^{\frac{1}{2}} \) gives \( (7(1) + 6)^{\frac{1}{2}} = (7 + 6)^{\frac{1}{2}} = \sqrt{13} \). Substituting \( x = 1 \) into \( (9 + 4x)^{\frac{1}{2}} \) gives \( (9 + 4(1))^{\frac{1}{2}} = (9 + 4)^{\frac{1}{2}} = \sqrt{13} \). Both sides of the original equation are equal when \( x = 1 \) so \( x = 1 \) is not an extraneous solution and is the final answer.

Key Concepts

Extraneous Solutions

When solving radical equations, sometimes you'll encounter solutions that don't actually satisfy the original equation. These are known as extraneous solutions. They can crop up when we perform operations like squaring both sides of an equation, which may introduce solutions that aren't initially valid.
To identify extraneous solutions, always substitute the solutions back into the original equation to verify them. If you find that both sides of your equation don't balance, that solution is extraneous. In this case, checking is crucial.
For our example, when we solved the equation \((7x + 6)^{\frac{1}{2}} = (9 + 4x)^{\frac{1}{2}}\), we squared both sides to eliminate the roots. After finding a potential solution, we carefully substituted it back in to show it was valid, ensuring it was not extraneous.

Solving Equations

Solving equations involves finding the value of the variable that makes the equation true. For radical equations, like the one in our exercise, the first step is often to eliminate the radical to simplify the equation.
We achieve this by performing algebraic operations, such as squaring both sides of a radical equation, to remove the square root. After this, it becomes much easier to solve for the variable using standard algebraic techniques.
In our problem, after squaring both sides gave us the equation \(7x + 6 = 9 + 4x\). We then used basic algebraic manipulation to solve for \(x\). This involved isolating \(x\) by moving the terms around, eventually finding that \(x = 1\).
Always remember, the key to solving any equation is applying operations systematically while ensuring you maintain the equality of both sides.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying expressions to uncover the value of unknowns. When dealing with equations, it's all about isolating the variable of interest through a series of steps.
In the context of our radical equation, we squared both sides to remove the radicals, resulting in a simplified algebraic equation. The goal was to isolate \(x\).
We subtracted \(4x\) from both sides to bring all terms involving \(x\) together, and then subtracted \(6\) from both sides to gather constant terms on one side. These steps gave us \(3x = 3\). Dividing by \(3\) finished the process, revealing \(x = 1\).
Effective algebraic manipulation not only makes solving equations easier but also helps ensure clarity at each step, preventing mistakes and confusion.