Problem 66 - Elementary Algebra

Question

For the following problems, solve the equations by completing the square or by using the quadratic formula. $$ a^{2}=a+4 $$

Step-by-Step Solution

Answer

Question: Solve the quadratic equation $$a^2 = a + 4$$ by completing the square and find the solution set for 'a'. Answer: The solution set for 'a' is $$a_1 = -\frac{1}{2} + \sqrt{\frac{15}{4}}$$ and $$a_2 = -\frac{1}{2} - \sqrt{\frac{15}{4}}$$.

Step 1: Write the equation in the standard quadratic form

To do this, we will subtract 'a' and add '-4' to both sides of the given equation. This will give us the equation in the standard form: $$a^2 - a - 4 = 0$$

Step 2: Complete the square

In order to complete the square, we will need to rewrite the equation in the form $$(a-h)^2=k$$, where 'h' and 'k' are the constants we need to determine. First, we will find the value of 'h', which is equal to $$\frac{1}{2}$$ times the linear coefficient of the equation, which is '-1' in our case. So, $$h=\frac{1}{2} \times (-1) = -\frac{1}{2}$$ Now, replace the linear term in the equation with h: $$a^2 - 2 \times a \times (-\frac{1}{2}) - (\frac{1}{2})^2 - 4 = 0$$ $$a^2 + a - \frac{1}{4} - 4 = 0$$ Add \((\frac{1}{4})\) to both sides of the equation to complete the square: $$a^2 + a - \frac{1}{4} = 4 - \frac{1}{4}$$ $$a^2 + a - \frac{1}{4} = \frac{15}{4}$$ $$\left( a + \frac{1}{2} \right)^2 = \frac{15}{4}$$

Step 3: Solve for 'a'

Now we can solve for 'a' by taking the square root of both sides of the equation: $$a + \frac{1}{2} = \pm \sqrt{\frac{15}{4}}$$ Subtract \(\frac{1}{2}\) from both sides to isolate 'a': $$a = -\frac{1}{2} \pm \sqrt{\frac{15}{4}}$$ The final solution set of the equation is: $$a_1 = -\frac{1}{2} + \sqrt{\frac{15}{4}}$$ $$a_2 = -\frac{1}{2} - \sqrt{\frac{15}{4}}$$

Key Concepts

Completing the Square

Completing the square is a method used to solve quadratic equations. This technique transforms a quadratic expression into a perfect square trinomial, which makes it easier to solve for the variable. Let's see how it works.

• Start with a quadratic equation in the form of \(a^2 + ba + c = 0\).
• Your goal is to rewrite this equation in the form of \((a - h)^2 = k\), where \(h\) and \(k\) are constants.

To find \(h\), use the formula \(h = \frac{1}{2} \text{ (linear coefficient)}\). For example, if the linear term is \(-a\), \(h\) would be \(-\frac{1}{2}\) since \(\frac{1}{2} \times (-1) = -\frac{1}{2}\). Next, complete the square:

• Replace the linear term with \(-2a\) times \(h\).
• Add \((h)^2\) to both sides of the equation to balance it.

This process will help you form a complete square that can be easily solved.

Standard Quadratic Form

The standard quadratic form is a way of writing quadratic equations. This form is written as \(as^2 + bs + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). Having the equation in this format is essential for solving it through different methods like completing the square or using the quadratic formula.
Let's convert a simple equation to the standard form.

• Take an equation like \(s^2 = s + 4\).
• To convert it to standard form, move all terms to one side, making the equation equal to zero.
• Subtract \(s\) and \(4\) from both sides: \(s^2 - s - 4 = 0\).

Now, it is in the standard form, ready to be solved.
Working with it in this way makes applying other techniques much easier.

Quadratic Formula

The quadratic formula is another powerful method for solving quadratic equations. It's especially useful when you can't easily factor the equation or don't want to complete the square. The formula is given by:\[a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Here’s how to apply it step by step:

• First, make sure the equation is in the standard form \(ax^2 + bx + c = 0\).
• Identify coefficients \(a\), \(b\), and \(c\).
• Substitute these values in the quadratic formula.
• Compute the discriminant \(b^2 - 4ac\). This tells you how many solutions to expect: if positive, there are two solutions; if zero, one solution; and if negative, no real solutions.

Using this method offers a straightforward way to find solutions, with all possible real solutions available depending on the discriminant.