Problem 68 - Elementary Algebra

Question

For the following problems, solve the equations by completing the square or by using the quadratic formula. $$ m^{2}-6=0 $$

Step-by-Step Solution

Answer

Question: Solve the quadratic equation \(m^2 - 6 = 0\) by completing the square or using the quadratic formula. Answer: The solutions for "m" are \(\pm\sqrt{6}\).

Step 1: Rewrite the equation in the form of \(ax^2 + bx + c = 0\)

The given equation is already in that form, with \(a = 1\), \(b = 0\), and \(c = -6\). $$ m^2 - 6 = 0 $$

Step 2: Complete the square for the equation

We want to find a value that we can add and subtract to both sides of the equation to make it a perfect square. To get this value, we can take half of the coefficient of \(m\) (which is 0 in this case) and square it. Since half of 0 is 0, and \(0^2 = 0\), this means that our equation is already a perfect square: $$ (m^2 - 6) + 0 = 0 + 0 $$

Step 3: Write the equation in square form and solve for "m"

Now that our equation is in the perfect square form, we can rewrite it as a square and solve for \(m\). In this case, we have: $$ (m^2 - 6) = 0 $$ We can rewrite the equation as: $$ (m^2 - 6) = 0 $$ Now we can solve for "m" by taking the square root of both sides: $$ \sqrt{m^2 - 6} = \sqrt{0} $$ Since the square root of 0 is 0: $$ m^2 - 6 = 0 $$ Now, we can solve for m by adding 6 to both sides of the equation: $$ m^2 = 6 $$ Finally, we take the square root of both sides of the equation to find our two possible solutions for "m": $$ m=\pm\sqrt{6} $$ So, the solutions for "m" are \(\pm\sqrt{6}\). Since we found the solutions through completing the square, there is no need to use the quadratic formula in this case.

Key Concepts

Completing the Square

Completing the square is a method used to solve quadratic equations by converting them into a perfect square trinomial. This approach can make solving these types of equations simpler and more intuitive. Here's how it works:

• Start by arranging the equation in the form \( ax^2 + bx + c = 0 \).
• Identify the coefficient of the linear term (the term with \(x\)), and take half of it. This value is critical for creating a perfect square.
• Square this halved coefficient and add and subtract it to the equation.

During this process, you often rewrite the quadratic expression into the form of \((x + d)^2\) or \((x - d)^2\). For instance, if we started with \( x^2 + 6x \):
- Half of 6 is 3, and squaring it gives us 9.- Adding and subtracting 9 allows us to write the expression as \((x + 3)^2 - 9\).Completing the square is extremely powerful as it allows for easy application of square root operations, leading you closer to finding the values of \(x\). In our example problem, the equation was already a perfect square, demonstrating how fast this method can find solutions.

Quadratic Formula

The quadratic formula is another approach to solving quadratic equations. It is especially useful when completing the square seems complicated or if coefficients are not easy to manipulate. The quadratic formula is derived from the standard quadratic equation \( ax^2 + bx + c = 0 \). It provides the solutions for \(x\) as follows:

\[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]- **\(b^2-4ac\)** is known as the discriminant and it determines the nature of the roots: - If the discriminant is positive, there are two real and distinct solutions. - If the discriminant is zero, there is one real solution. - If the discriminant is negative, the solutions are complex (not real numbers).

Using this formula ensures you always get correct solutions, regardless of whether the quadratic is a perfect square or not. In our exercise, attempting to use the quadratic formula might also provide the same roots \( \pm\sqrt{6} \), reaffirming your calculations from completing the square.

Perfect Square Form

A perfect square form of a quadratic equation is one where the equation can be expressed as the square of a binomial. Recognizing a perfect square trinomial within a quadratic equation can make solving it a lot easier. For example, if you have an expression like \( (x - 3)^2 = 9 \), it can instantly be solved by extracting the square root.

• Detecting a perfect square means identifying expressions of the form \((a + b)^2\) or \((a - b)^2\).
• Expand them to see if they match \( a^2 + 2ab + b^2 \) (or minus in the latter case).

For our original equation, because we ended up with \( m^2 - 6 = 0 \) and spotted that it doesn't fit into this expanded form exactly, we adjust only when needed.
The beauty of expressing a quadratic as a perfect square form is it allows immediate application of square roots, and in mathematical terms, reduces a quadratic equation problem to finding a linear solution. In circumstances where it's applicable without creating complicated fractions, it’s ideal to use.