Problem 79 - Algebra A Combined Function

Question

Use the quadratic formula and a calculator to approximate each solution to the nearest tenth. $$ 2 x^{2}-6 x+3=0 $$

Step-by-Step Solution

Answer

The solutions are approximately 2.4 and 0.6.

Step 1: Identify the coefficients

To solve the quadratic equation using the quadratic formula, start by identifying the coefficients in the equation. The given equation is \(2x^2 - 6x + 3 = 0\). Thus, we have \(a = 2\), \(b = -6\), and \(c = 3\).

Step 2: Write down the quadratic formula

The quadratic formula is used to find the solutions of a quadratic equation and is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Step 3: Calculate the discriminant

Calculate the discriminant \(b^2 - 4ac\). For our equation, it is:\((-6)^2 - 4(2)(3) = 36 - 24 = 12\).

Step 4: Plug the values into the quadratic formula

Substitute \(a\), \(b\), and \(c\) into the quadratic formula, using the discriminant calculated:\[x = \frac{-(-6) \pm \sqrt{12}}{2(2)}\]This simplifies to:\[x = \frac{6 \pm \sqrt{12}}{4}\].

Step 5: Simplify \(\sqrt{12}\)

\(\sqrt{12}\) can be simplified to \(2\sqrt{3}\). Thus the expression becomes:\[x = \frac{6 \pm 2\sqrt{3}}{4}\].

Step 6: Separate into two solutions

Calculate each solution separately:1. \(x_1 = \frac{6 + 2\sqrt{3}}{4}\)2. \(x_2 = \frac{6 - 2\sqrt{3}}{4}\).

Step 7: Approximate the solutions using a calculator

Use a calculator to find the approximate decimal values:1. \(x_1 \approx \frac{6 + 3.464}{4} = \frac{9.464}{4} \approx 2.4\) (nearest tenth)2. \(x_2 \approx \frac{6 - 3.464}{4} = \frac{2.536}{4} \approx 0.6\) (nearest tenth).

Key Concepts

Understanding the Discriminant

The discriminant is a key component when working with the quadratic formula. It is found within the square root symbol and is represented by the expression \(b^2 - 4ac\). The discriminant helps you determine the nature of the roots of the quadratic equation.
Here’s what the discriminant can tell you:

• If the discriminant is positive, the quadratic equation has two distinct real solutions.
• If it is zero, there is exactly one real solution.
• If the discriminant is negative, no real solutions exist, but two complex solutions are possible.

In the example equation \(2x^2 - 6x + 3 = 0\), the discriminant is calculated as \((-6)^2 - 4(2)(3) = 36 - 24 = 12\). Since 12 is positive, we know there are two distinct real solutions for this quadratic equation.

How to Solve Quadratic Equations

Solving quadratic equations involves finding the values of \(x\) that satisfy the equation. The quadratic formula is one reliable method for this. The formula is\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Using this formula tackles any quadratic equation of the form \(ax^2 + bx + c = 0\).
To use the formula successfully:

• Start by identifying the values of \(a\), \(b\), and \(c\) from your equation.
• Calculate the discriminant \(b^2 - 4ac\).
• Substitute these values into the quadratic formula.

For our specific equation \(2x^2 - 6x + 3 = 0\), we have \(a = 2\), \(b = -6\), and \(c = 3\). Plug these into the formula to get the solutions for \(x\).

Approximating Solutions

After finding the exact form of the solutions using the quadratic formula, it's often necessary to approximate these solutions, particularly when they involve irrational numbers. This is typically done using a calculator.
In cases like our problem, where we simplify \(\sqrt{12}\) to \(2\sqrt{3}\), you then need to calculate what \(2\sqrt{3}\) is approximately—specifically, \(\sqrt{3} \approx 1.732\), so \(2\sqrt{3} \approx 3.464\).
Once you have this approximation:

• Calculate the solutions \(x_1 = \frac{6 + 3.464}{4}\) and \(x_2 = \frac{6 - 3.464}{4}\).
• Using a calculator, you find the approximate decimal values which determine \(x_1 \approx 2.4\) and \(x_2 \approx 0.6\), rounded to the nearest tenth.

Approximating gives practical solutions applicable in realistic scenarios where exactness may not be vital.