Problem 2 - Introductory Algebra for College Students

Question

Solve each equation using the quadratic formula. Simplify irrational solutions, if possible. $$x^{2}+7 x+10=0$$

Step-by-Step Solution

Answer

The roots of the equation \(x^2 + 7x + 10 = 0\) are -2 and -5.

Step 1: Identify terms with the general quadratic formula

Given the general quadratic equation \(ax^2 + bx + c = 0\), we can identify that: \(a = 1, b = 7, c = 10\) for the equation in the problem, that is \(x^2 + 7x + 10 = 0\).

Step 2: Substitute the values into the quadratic formula and simplify

The quadratic formula is \(\frac{-b \pm \sqrt{b^2 -4ac}}{2a}\). Substituting \(a = 1, b = 7, c = 10\), we get \(\frac{-7 \pm \sqrt{7^2 - 4*1*10}}{2*1}\) which simplifies to \(\frac{-7 \pm \sqrt{49 - 40}}{2}\). This further simplifies to \(\frac{-7 \pm \sqrt{9}}{2}\).

Step 3: Calculate the roots

Considering the plus and minus signs separately, we find the two roots of the equation. So the roots are \(\frac{-7 + 3}{2} = -2\) and \(\frac{-7 - 3}{2} = -5\).

Key Concepts

Solving Quadratic Equations

Quadratic equations often stand in the form of \(ax^2 + bx + c = 0\). Solving these equations means finding the values of \(x\) that make the equation true. One of the most widely used methods to solve quadratic equations is the quadratic formula. This formula is a tool, ready to complete the task for us. To start, you first need to recognize the coefficients and constants in the given equation. In our example, the equation is \(x^2 + 7x + 10 = 0\), where:

• \(a =1\): the coefficient in front of \(x^2\)
• \(b=7\): the coefficient in front of \(x\)
• \(c=10\): the constant term

Once you have these values, you substitute them into the quadratic formula. Applying this tool with consistent practice will make problem-solving easier and faster.

Step-by-Step Solution

Let's see how to put the quadratic formula into action, step by step. This formula is \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Once you've identified your values of \(a\), \(b\), and \(c\), the next step is substitution.
Here's how it unfolds:

• Substitute to get: \(\frac{-7 \pm \sqrt{7^2 - 4 \times 1 \times 10}}{2 \times 1}\)
• Calculate inside the square root: \(\sqrt{49 - 40}\)
• This becomes \(\sqrt{9}\), simplifying to 3

Now, using both plus and minus (\(\pm\)) in the formula:

• \(\frac{-7 + 3}{2} = -2\)
• \(\frac{-7 - 3}{2} = -5\)

We have solved for two roots: \(x = -2\) and \(x = -5\). Practicing these steps is essential to becoming confident in solving any quadratic equation.

Simplifying Irrational Solutions

Sometimes, the roots of quadratic equations can be irrational. Irrational numbers are those that cannot be expressed as a simple fraction, often leading to a non-terminating decimal. For example, solutions involving \(\sqrt{3}\) or \(\sqrt{5}\). At times, however, like in our equation, solutions are rational and can be simplified easily.In such cases, the square root in the quadratic formula simplifies to a whole number. As seen here, \(\sqrt{9} = 3\), leading to straightforward results.

• If simplifying is possible, do not hesitate to express the roots in simplest form.
• If roots were irrational like \(\sqrt{2}\), leave them as they are or round off if needed.

Simplifying solutions makes them easier to interpret and use in further calculations. Be watchful whether \'b²-4ac' leads to a perfect square or not, as this determines the nature of your roots.