Problem 38 - Algebra 1

Question

Solve the equation. Check for extraneous solutions. $$x=\sqrt{6 x-9}$$

Step-by-Step Solution

Answer

The solution to the equation is \(x = 3\).

Step 1: Square both sides

In this step, square both sides of the equation to eliminate the square root. The equation becomes: \(x^2 = (6x-9)\}. Now simplify the right hand side to get \(x^2 - 6x + 9 = 0\)

Step 2: Solve the quadratic equation

The equation in step 1 is a quadratic equation. Solve it by factorising, thereby obtaining two potential solutions for \(x\). For \(x = 3\), check them against the original equation in order to confirm whether they are valid solutions.

Step 3: Check for extraneous solutions

After calculating the roots as \(x = 3\), substitute them back into the original equation and verify whether the left-hand side equals the right-hand side. For \(x = 3\), it does, so there are no extraneous solutions in this case.

Key Concepts

Extraneous Solutions

Extraneous solutions often appear when solving equations involving square roots. This happens when we manipulate the equation, like squaring both sides, which can sometimes introduce solutions that don't satisfy the original equation.
To detect these solutions, always substitute your results back into the initial expression and check if the equation holds true. For the exercise given, when we square both sides to remove the square root, new potential solutions may emerge. Thus, after solving the quadratic, substitute each solution into the original equation:

• If the substituted solution satisfies the equation, it is valid.
• If not, it's an extraneous solution.

In our example, after solving the quadratic, we found that only one solution, \(x = 3\), satisfies the original equation. Therefore, there were no extraneous solutions.

Factorization

Factorizing is a critical skill when dealing with quadratic equations, like the one we handled in this exercise.
To factor a quadratic equation such as \[x^2 - 6x + 9 = 0\],seek two numbers whose product is the constant term (which is 9) and whose sum is the coefficient of the linear term (-6).

• In this case, the numbers -3 and -3 satisfy both conditions, giving us the factorized form \((x - 3)(x - 3) = 0\).

This method is efficient and often brings out solutions faster by breaking down complex equations into more manageable pieces.
Once factorized, setting each factor equal to zero provides potential solutions for the original quadratic, which you then validate.

Square Root Property

The square root property is used to solve equations where the variable appears inside a square root symbol.
By squaring both sides of the equation, we eliminate the square root, transforming the expression into a quadratic equation that can be managed using standard techniques such as factorization.

• For example, in the equation \(x = \sqrt{6x - 9}\), squaring both sides results in \(x^2 = 6x - 9\).
• This process allows us to draw out a polynomial equation, which we can then handle with our algebraic toolbox.

However, remember that this step can introduce extraneous solutions, requiring additional verification.
It is crucial to approach solving square root equations with care, ensuring all potential solutions are checked against the original equation.