Problem 32 - Elementary Algebra

Question

For the following problems, solve the square root equations. $$ \sqrt{x-7}-\sqrt{5 x+1}=0 $$

Step-by-Step Solution

Answer

Answer: There is no solution to this equation as the value obtained for x results in a negative value inside the square roots.

Step 1: Isolate one of the square roots

To isolate one of the square roots, we will add the second square root to both sides of the equation: $$ \sqrt{x-7} = \sqrt{5x + 1} $$

Step 2: Square both sides of the equation

To eliminate the square roots, square both sides of the equation: $$ (\sqrt{x-7})^2 = (\sqrt{5x + 1})^2 $$ This simplifies to: $$ x - 7 = 5x + 1 $$

Step 3: Simplify the equation

Subtract 'x' from both sides and subtract '1' from both sides to get: $$ -7-x+1 = 5x-x $$ Now simplify it to: $$ -6 = 4x $$

Step 4: Solve for x

Divide both sides by 4 to isolate 'x': $$ x = -\frac{6}{4} $$ Simplify the fraction to get: $$ x = -\frac{3}{2} $$

Step 5: Check the solution

We need to make sure our solution is valid. Plug the value of x back into the original equation: $$ \sqrt{-\frac{3}{2}-7}-\sqrt{5\left(-\frac{3}{2}\right)+1}=0 $$ Simplify inside the square roots: $$ \sqrt{-\frac{17}{2}}-\sqrt{-\frac{7}{2}}=0 $$ Since we cannot have a negative value inside a square root, our solution is not valid. Therefore, there is no solution to this equation.

Key Concepts

Equation Solving

Solving equations, especially square root equations, involves finding the value of the variable that satisfies the equation. In the given problem, we aim to solve the equation \( \sqrt{x-7} - \sqrt{5x+1} = 0 \). The goal is to find if there exists a value for \( x \) that makes the expression equal to zero.
Here is a simple way to approach solving square root equations:

• Ensure that both sides of the equation are simplified.
• Remove the square roots by squaring both sides, if necessary.
• Simplify further to reach a linear or simpler equation.
• Finally, check whether the solution truly satisfies the original equation.

This technique ensures that every step brings you closer to solving the equation effectively.

Isolating Square Roots

Isolating the square root is a crucial step in solving equations that involve radicals. In our example, to handle \( \sqrt{x-7} - \sqrt{5x+1} = 0 \), we need to isolate one or both of the square roots.
To isolate, perform the following:

• Move one square root to the opposite side of the equation. In this case, we will add \( \sqrt{5x+1} \) to both sides, resulting in \( \sqrt{x-7} = \sqrt{5x+1} \).

This forms a more straightforward basis for squaring, which will help eliminate the square roots altogether. However, be careful while isolating radicals, as it can introduce extraneous solutions. Always isolate the simplest form before proceeding.

Checking Solutions

After solving for the variable, it is crucial to check the solution in the original equation. This step is key because squaring both sides can sometimes introduce solutions that don't actually satisfy the initial conditions.
For our problem:

• We found \( x = -\frac{3}{2} \) after solving.
• Substitute \( x \) back into the original equation: \( \sqrt{-\frac{3}{2}-7} - \sqrt{5(-\frac{3}{2})+1} \).
• Calculate the expressions: \( \sqrt{-\frac{17}{2}} - \sqrt{-\frac{7}{2}} = 0 \).

Realizing that you cannot have negatives inside square roots, this indicates our solution does not satisfy the original equation. Thus, concluding no valid solution exists. Always perform this verification to avoid assuming an incorrect solution.

Simplifying Fractions

Simplifying fractions can make solving equations more manageable and results clearer. Through the steps, we simplified \( -\frac{6}{4} \) to \( -\frac{3}{2} \), making it cleaner and easier to interpret.When simplifying fractions, remember:

• Divide the numerator and the denominator by their greatest common divisor (GCD).
• This results in a fraction that is simple and reduced to its lowest terms.

Simplification is always helpful before plugging values back into the equation, as it makes calculations easier and reduces error chances. It aligns the solution to how it's typically presented in standard mathematical form.