Problem 114 - Elementary Algebra

Question

Solve by using the Quadratic Formula. \(3 w(w-2)-8=0\)

Step-by-Step Solution

Answer

w = 1 + \( \frac{\sqrt{33}}{3} \) or w = 1 - \( \frac{\sqrt{33}}{3} \)

Step 1: Expand the Expression

First, expand the expression on the left-hand side: \[ 3 w(w-2) - 8 = 0 \] Multiply out the terms inside the parentheses: \[ 3w^2 - 6w - 8 =0 \]

Step 2: Identify Coefficients

Identify the coefficients in the quadratic equation in the standard form \( ax^2 + bx + c = 0 \). Here, \( a = 3 \), \( b = -6 \), and \( c = -8 \).

Step 3: Apply the Quadratic Formula

The quadratic formula is given by: \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute the identified coefficients into the formula: \[ w = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 3 \cdot (-8)}}{2 \cdot 3} \]This simplifies to: \[ w = \frac{6 \pm \sqrt{36 + 96}}{6} \]

Step 4: Simplify the Radicand

Simplify the expression under the square root: \[ w = \frac{6 \pm \sqrt{132}}{6} \]Further simplification of the square root gives: \[ w = \frac{6 \pm 2\sqrt{33}}{6} \]

Step 5: Simplify the Expression

Divide each term by 6: \[ w = 1 \pm \frac{\sqrt{33}}{3} \]

Step 6: Write the Solution

The solutions to the quadratic equation are: \[ w = 1 + \frac{\sqrt{33}}{3} \] and \[ w = 1 - \frac{\sqrt{33}}{3} \]

Key Concepts

expanding expressions

When solving quadratic equations, it's important to understand how to expand expressions. In our exercise, we start with the equation \(3 w(w-2)-8=0\).

We need to get rid of the parentheses. To do this, distribute \(3w\) across each term inside the parentheses: \(3w \times w\text{ and } 3w \times -2\).

This results in the expanded form: \(3w^2 - 6w - 8 = 0\).
The equation is now in the standard form needed to apply the quadratic formula.
This process of expanding expressions makes the equation solvable. Remember: Distribute carefully and double-check all multiplications.

identifying coefficients

After expanding the expression, the next critical step is identifying the coefficients. The standard form for a quadratic equation is: \(ax^2 + bx + c = 0\).

In our expanded equation, \(3w^2 - 6w - 8 = 0\), we can see that:

• \(a = 3\)
• \(b = -6\)
• \(c = -8\)

Identifying these coefficients correctly is essential because they are inserted into the quadratic formula. If the coefficients are identified wrong, it will lead to incorrect solutions. Always ensure to recognize and extract the coefficients with the right signs.

solving quadratic equations

Once coefficients are identified, we can proceed to solve the quadratic equation using the quadratic formula: \[w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].

Plugging in our coefficients (\(a = 3\), \(b = -6\), \(c = -8\)):

This gives us: \[w = \frac{-(-6) \pm \sqrt{ (-6)^2 - 4 \cdot 3 \cdot (-8)}}{2 \cdot 3}\].

simplify this to:
\[w = \frac{6 \pm \sqrt{36 + 96}}{6}\].
However, solving further requires simplifying the expression under the square root (the radicand). This careful substitution and simplification lead us closer to finding \(w\).

simplifying radicals

In the quadratic formula, we'll often encounter square roots (radicals). Simplifying these radicals is crucial. With our expression \(\sqrt{36 + 96}\), we simplify inside the square root first: \(36 + 96 = 132\).

So our problem becomes: \(\sqrt{132}\).

Breaking this down: \(132 = 4 \times 33\). Since we can take the square root of 4, substitute that in to get:

\(\sqrt{4 \times 33} = 2 \sqrt{33}\).

Then, proceeding with our quadratic formula, we have:
\[w = \frac{6 \pm 2 \sqrt{33}}{6}\].

Further simplification gives us the solution: \(w = 1 + \frac{\sqrt{33}}{3}\) and \(w = 1 - \frac{\sqrt{33}}{3}\). Remember, simplifying radicals helps us get the final, most precise solution.