Problem 16 - Algebra 2 and Trigonometry

Question

In \(3-14\) , use the quadratic formula to find the roots of each equation. Irrational roots should be written in simplest radical form. $$ 4 x^{2}-4 x=11 $$

Step-by-Step Solution

Answer

The roots are \(x = \frac{1 + 2\sqrt{3}}{2}\) and \(x = \frac{1 - 2\sqrt{3}}{2}\).

Step 1: Write the equation in standard form

To use the quadratic formula, the equation needs to be in the standard form: \(ax^2 + bx + c = 0\). The given equation is \(4x^2 - 4x = 11\). Begin by subtracting 11 from both sides to get \(4x^2 - 4x - 11 = 0\).

Step 2: Identify coefficients

Identify the coefficients \(a\), \(b\), and \(c\) from the standard form equation \(4x^2 - 4x - 11 = 0\). Here, \(a = 4\), \(b = -4\), and \(c = -11\).

Step 3: Write down the quadratic formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). We'll use this formula to find the roots of the equation.

Step 4: Calculate the discriminant

Compute the discriminant: \(b^2 - 4ac\). For our equation, this is \((-4)^2 - 4 \times 4 \times (-11) = 16 + 176 = 192\). Since the discriminant is positive, there will be two real roots.

Step 5: Simplify and solve using the quadratic formula

Substitute \(a\), \(b\), and \(c\) into the quadratic formula: \(x = \frac{-(-4) \pm \sqrt{192}}{2 \times 4}\). This simplifies to \(x = \frac{4 \pm \sqrt{192}}{8}\).

Step 6: Simplify \(\sqrt{192}\)

Simplify \(\sqrt{192}\) by finding its prime factorization: \(192 = 2^6 \times 3\). So \(\sqrt{192} = 2^3 \times \sqrt{3} = 8\sqrt{3}\).

Step 7: Finalize the roots

Substitute \(\sqrt{192} = 8\sqrt{3}\) back into the formula: \(x = \frac{4 \pm 8\sqrt{3}}{8}\). Simplify to get the roots: \(x = \frac{1 \pm 2\sqrt{3}}{2}\).

Key Concepts

Understanding Irrational Roots

Irrational roots can be a bit mind-bending at first, but we've got you covered. These are roots or solutions of equations that cannot be expressed as simple fractions or integers. Instead, they are usually expressed in terms of square roots, cube roots, or higher-order roots that seem "messy" or "complicated." For example, in our solution, we ended up with roots involving \( \sqrt{3} \).

This isn't peculiar; it's a natural part of working with quadratic equations that don't have perfect squares within the discriminant. When you break down numbers like \( \sqrt{192} \) into simplest radical form—simplifying it to \( 8\sqrt{3} \) in our case—you are revealing its irrational nature. Keep in mind that:

• Irrational roots emerge when the discriminant (more on this below) is not a perfect square.
• These roots add an extra layer of beauty and complexity to algebra.
• You will often leave them in radical form unless instructed to provide a decimal approximation.

Decoding the Discriminant

The discriminant may sound complex, but it's simply a part of the quadratic formula used to help you understand the nature of the roots. It's given by \( b^2 - 4ac \). This nifty term tells us quite a bit about the solutions to the quadratic equation in question.

Here's how it works:

• If the discriminant is positive, like in our example where \( b^2 - 4ac = 192 \), you'll get two real roots.
• If it's zero, you have one real root—a so-called "double root."
• If negative, it indicates that the roots are complex or imaginary (not "real" numbers).

Understanding whether a discriminant is a perfect square can also tell you whether those roots are rational (expressed as a fraction) or irrational (involving a radical). So always calculate the discriminant first to get a sneak peek into the results before diving into solving the equation fully.

Prime Factorization Simplifies Radicals

Prime factorization is our secret weapon for simplifying radicals, especially when dealing with large square roots. To do this, you break down the number into its basic building blocks—its prime factors.

For example, in our exercise, we had \( \sqrt{192} \). By identifying the prime factors of 192 as \( 2^6 \times 3 \), we were able to simplify the square root considerably:

• Identify that \( 192 = 2^6 \times 3 \).
• Recognize that the square root of \( 2^6 \) is \( 2^3 = 8 \).
• The square root \( \sqrt{192} \) thus becomes \( 8\sqrt{3} \).

Prime factorization lets us write radicals in their simplest form, reducing complexity and making further calculations much easier. Getting comfortable with this skill can certainly make working with radicals and quadratic equations far less daunting.