Problem 98 - Elementary Algebra

Question

For exercises 97-100, the completed problem has one mistake. (a) Describe the mistake in words, or copy down the whole problem and highlight or circle the mistake. (b) Do the problem correctly. Problem: Solve: \(2 x^{2}-12 x-54=0\) Incorrect Answer: \(2 x^{2}-12 x-54=0\) \(2\left(x^{2}-6 x-27\right)=0\) \(2(x-9)(x+3)=0\) \(2=0\) or \(x-9=0\) or \(x+3=0\) \(\frac{+9+9}{x=9}\) or \(\frac{-3-3}{x=-3}\) Solutions: \(x=2, x=9, x=-3\)

Step-by-Step Solution

Answer

The correct solutions are \(x = 9\) and \(x = -3\).

Step 1: Identify the Mistake

The mistake is in factoring the equation incorrectly. The step where the incorrect solution shows \[2(x-9)(x+3)=0\] is wrong. Therefore, the subsequent steps and the solutions derived from that factored form are incorrect.

Step 2: Start with the Given Equation

Begin with the original quadratic equation: \[2x^{2} - 12x - 54 = 0\].

Step 3: Factor Out the Common Term

Factor out the common factor, which is 2, from the quadratic equation: \[2(x^{2} - 6x - 27) = 0\].

Step 4: Simplify the Equation

Divide both sides of the equation by 2:\[x^{2} - 6x - 27 = 0\].

Step 5: Factor the Quadratic Equation

Find two numbers that multiply to -27 and add to -6. These numbers are -9 and +3. Therefore, the factored form is:\[(x - 9)(x + 3) = 0\].

Step 6: Solve for x

Set each factor equal to zero and solve for x:\[x - 9 = 0\] gives \[x = 9\], and\[x + 3 = 0\] gives \[x = -3\].

Key Concepts

Factoring Quadratics

Factoring quadratics is a key method to solve quadratic equations. The general form of a quadratic equation is given by \(ax^2 + bx + c = 0\). To factor this, we need to find two numbers that both add up to \(b\) and multiply to \(ac\).

For example, in the equation \(2x^2 - 12x - 54 = 0\), we first simplify by factoring out the common factor of 2: \[2(x^2 - 6x - 27) = 0 \] Next, we need to find two numbers that multiply to -27 and add to -6; these numbers are -9 and +3. This lets us write the equation as: \[(x - 9)(x + 3) = 0\].

Each factor represents a potential solution. Solving these smaller equations gives the values for x that solve the original quadratic equation.

Identifying Mistakes

Identifying mistakes in solving quadratics is crucial to ensure that we arrive at the correct solutions. Often, errors arise in the factoring or simplification steps.
For instance, in the incorrect solution provided for the problem \(2x^2 - 12x - 54 = 0\), the mistake is made during factoring. The incorrect step given is: \[2(x-9)(x+3)=0\]
However, this factorization is incorrect as it does not properly split the middle term.
The correct factoring should be: \[(x-9)(x+3) = 0\].
Spotting this mistake early prevents wrong solutions and saves time.

Steps to Solve Quadratics

There is a straightforward process to solve quadratic equations:

• Start with the given quadratic equation.
• Simplify by factoring out any common terms.
• Factor the quadratic expression.
• Set each factor equal to zero and solve for x.

Let's apply these steps to solve \(2x^2 - 12x - 54=0\):

1. Original equation: \[2x^2 - 12x - 54 = 0 \]
2. Factor out the common term: \[2(x^2 - 6x - 27) = 0\]
3. Divide both sides by 2: \[x^2 - 6x - 27 = 0 \]
4. Find factors: \[(x-9)(x+3)=0\]
5. Solve for x: \[x - 9 = 0 \rightarrow x = 9 \] and \[x + 3 = 0 \rightarrow x = -3 \]

Common Factors in Equations

Identifying common factors in quadratic equations can simplify the process of solving them. This means looking for a number or variable that can be factored out from all terms in the equation.

For instance, in the quadratic \[2x^2 - 12x - 54 = 0\], all terms share a common factor of 2. Factoring out the 2, we get: \[2(x^2 - 6x - 27) = 0\]
The equation is now easier to work with. Always check for common factors before proceeding with more complex factoring. This can save time and reduce mistakes, streamlining the equation-solving process.