Problem 31 - Algebra 1

Question

Solve the equation. Check for extraneous solutions. $$\sqrt{\frac{1}{5} x-2}-\frac{1}{10}=\frac{7}{10}$$

Step-by-Step Solution

Answer

The solution to the equation is \(x = 13.2\). After checking, there are no extraneous solutions.

Step 1: Isolate the square root expression

To start off, the equation needs to be simplified by isolating the square root expression. This can be achieved by adding \(\frac{1}{10}\) to both sides of the equation. The equation is then: \[ \sqrt{\frac{1}{5} x-2} = \frac{8}{10} \] which simplifies further to: \[ \sqrt{\frac{1}{5} x-2} = \frac{4}{5} \]

Step 2: Square both sides of the equation

To eliminate the square root, square both sides of the equation:\[ (\sqrt{\frac{1}{5}x - 2})^2 = (\frac{4}{5})^2 \] This simplifies to: \[ \frac{1}{5}x - 2 = \frac{16}{25} \]

Step 3: Solve for x

Rearrange the equation to get x by itself. Addition of 2 to both sides yields: \[ \frac{1}{5}x = \frac{16}{25} + 2 \] which simplifies to: \[ \frac{1}{5}x = \frac{16}{25} + \frac{50}{25} \] further simplifies to: \[ \frac{1}{5}x = \frac{66}{25} \] and by multiplying both sides by 5, the solution to x is: \[ x = \frac{66}{5} = 13.2\]

Step 4: Check for extraneous solutions

Substitute x = 13.2 into the original equation to check if it's a valid solution:\[ \sqrt{\frac{1}{5} * 13.2 - 2} - \frac{1}{10} = \frac{7}{10} \] which simplifies to \[ \frac{8}{10} - \frac{1}{10} = \frac{7}{10} \] Since both sides are equal, x = 13.2 is a valid solution and there are no extraneous solutions.

Key Concepts

Extraneous Solutions

In mathematics, extraneous solutions are solutions derived during the process of solving that are not valid when plugged back into the original equation. These can appear especially when dealing with equations involving square roots and other higher-degree polynomials. It's vital to check every potential solution against the original equation to ensure its validity.
This check serves to confirm that the solution satisfies all given conditions and constraints.
For our original equation \( \sqrt{\frac{1}{5}x - 2} - \frac{1}{10} = \frac{7}{10} \), once we arrived at the solution \( x = 13.2 \), we substituted this value back into the original equation.
When substituting, if the right-hand side equals the left-hand side of the equation, the solution is confirmed. If not, the solution is extraneous and should be discarded. In this particular exercise, replacing \( x \) with \( 13.2 \) yields equal results on both sides of the equation, proving that there are no extraneous solutions here.

Isolating Square Root

To solve equations involving square roots, the first step often involves isolating the square root term.
This means rearranging the equation so the square root expression stands alone on one side.
By doing so, it becomes feasible to apply further operations, like squaring, without affecting other terms.
In the given exercise, we started by keeping \( \sqrt{\frac{1}{5}x - 2} \) alone on one side by adding \( \frac{1}{10} \) to both sides of the equation. We transformed the equation from \( \sqrt{\frac{1}{5} x-2}-\frac{1}{10} = \frac{7}{10} \) into a simpler form \( \sqrt{\frac{1}{5} x-2} = \frac{4}{5} \).
It's the clean separation of terms that allows us to proceed to the next methodical stage: squaring.

Squaring Both Sides

Once the square root term is isolated, the next logical step in solving the equation is to square both sides.
This is the process of raising both sides of the equation to the power of two, which effectively removes the square root.
For the formula \( \sqrt{\frac{1}{5}x - 2} = \frac{4}{5} \), squaring this results in \( \left(\sqrt{\frac{1}{5}x - 2}\right)^2 = \left(\frac{4}{5}\right)^2 \).
The square root and the square cancel each other on the left side, simplifying to \( \frac{1}{5}x - 2 \). On the right side, \( \left(\frac{4}{5}\right)^2 \) becomes \( \frac{16}{25} \).
This transformation is crucial because it allows us to then solve the equation algebraically, ultimately solving for \( x \).
By squaring, we facilitate an easier calculation, which is often a crucial step towards finding meaningful solutions.