Problem 44 - Algebra 2

Question

Solve each equation by using the method of your choice. Find exact solutions. \(2 x^{2}-12 x+7=5\)

Step-by-Step Solution

Answer

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

Step 1: Simplify the Equation

Subtract 5 from both sides of the equation to bring it to standard quadratic form: \(2x^2 - 12x + 7 - 5 = 0\). This simplifies to \(2x^2 - 12x + 2 = 0\).

Step 2: Divide the Entire Equation

Divide each term by 2 to simplify the equation: \(x^2 - 6x + 1 = 0\).

Step 3: Use the Quadratic Formula

Recall the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). For our equation \(x^2 - 6x + 1 = 0\), \(a = 1\), \(b = -6\), and \(c = 1\).

Step 4: Calculate the Discriminant

Calculate \(b^2 - 4ac\): \((-6)^2 - 4 \times 1 \times 1 = 36 - 4 = 32\).

Step 5: Find the Solutions

Plug the values into the quadratic formula: \[x = \frac{-(-6) \pm \sqrt{32}}{2 \times 1} = \frac{6 \pm \sqrt{32}}{2}\]. Simplify \(\sqrt{32}\) to \(4\sqrt{2}\), giving: \[x = \frac{6 \pm 4\sqrt{2}}{2}\].

Step 6: Simplify the Expression

Further simplify by dividing both terms by 2: \(x = 3 \pm 2\sqrt{2}\).

Key Concepts

Quadratic Formula

The quadratic formula is a key tool for solving quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). When solving any quadratic equation, the quadratic formula allows us to find the values of \(x\) by substituting \(a\), \(b\), and \(c\) from the equation into:
- \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
This formula is particularly useful when factoring is complex or impossible. To apply it, you must know:

• \(a\) represents the coefficient of \(x^2\).
• \(b\) is the coefficient of \(x\).
• \(c\) is the constant term.

In our problem, after simplifying and dividing by 2, our quadratic equation became \(x^2 - 6x + 1 = 0\). Here, \(a = 1\), \(b = -6\), and \(c = 1\). By substituting these values into the formula, we can find the exact solutions.

Discriminant Calculation

The discriminant is an essential part of the quadratic formula, contained within the square root \(\sqrt{b^2 - 4ac}\). Calculating the discriminant helps determine the number and type of solutions a quadratic equation has.
Here’s how it works:

• If the discriminant (\(b^2 - 4ac\)) is positive, the equation has two distinct real solutions.
• If it is zero, there is exactly one real solution (repeated root).
• If it is negative, the equation has two complex solutions.

For the equation \(x^2 - 6x + 1 = 0\), the discriminant is calculated as follows:\((-6)^2 - 4 \times 1 \times 1 = 36 - 4 = 32.\)Since our discriminant is 32, which is positive, this indicates two distinct real solutions.

Exact Solutions

Finding exact solutions involves plugging values into the quadratic formula and simplifying them appropriately. In our example, using the quadratic formula, we calculated:\(x = \frac{-(-6) \pm \sqrt{32}}{2 \times 1} = \frac{6 \pm \sqrt{32}}{2}.\)Next, the square root \(\sqrt{32}\) simplifies to \(4\sqrt{2}\). Substituting this in the formula, we get:\[x = \frac{6 \pm 4\sqrt{2}}{2}.\]Finally, simplify this expression by dividing the terms in the numerator by 2:\[x = 3 \pm 2\sqrt{2}.\]This renders our exact solutions \(x = 3 + 2\sqrt{2}\) and \(x = 3 - 2\sqrt{2}\). Each solution is exact and expressed using roots, retaining precision crucial for higher-level mathematics or applications.