Problem 109 - Elementary Algebra

Question

Solve by using the Quadratic Formula. \(2 a^{2}-6 a+3=0\)

Step-by-Step Solution

Answer

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

Step 1 - Identify the coefficients

Identify the coefficients in the quadratic equation \(2a^2 - 6a + 3 = 0\). Here, the coefficient of \(a^2\) is \(a = 2\), the coefficient of \(a\) is \(b = -6\), and the constant term is \(c = 3\).

Step 2 - Write down the Quadratic Formula

The Quadratic Formula is given by: \[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Plug in the values of \(a\), \(b\), and \(c\).

Step 3 - Calculate the discriminant

Calculate the discriminant using the formula: \(b^2 - 4ac\). Substituting the given values: \((-6)^2 - 4(2)(3) = 36 - 24 = 12\).

Step 4 - Find the square root of the discriminant

Calculate the square root of the discriminant: \(\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}\).

Step 5 - Substitute and simplify

Substitute \(\sqrt{12}\) back into the Quadratic Formula: \[ a = \frac{-(-6) \pm 2\sqrt{3}}{2(2)} \] Simplify: \[ a = \frac{6 \pm 2\sqrt{3}}{4} = \frac{6}{4} \pm \frac{2\sqrt{3}}{4} = \frac{3}{2} \pm \frac{\sqrt{3}}{2} \].

Step 6 - State the solutions

The solutions to the quadratic equation are: \[ a = \frac{3}{2} + \frac{\sqrt{3}}{2} \] and \[ a = \frac{3}{2} - \frac{\sqrt{3}}{2} \].

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation involving a variable raised to the second power. It generally has the form: ax^2 + bx + c = 0, where a, b, and c are coefficients, and a ≠ 0. Quadratic equations are essential because they appear in various mathematical contexts, and understanding how to solve them is a key algebraic skill. The quadratic term (ax^2), the linear term (bx), and the constant term (c) together shape the parabola described by the equation when graphed.
Let's solve the example given: 2a^2 - 6a + 3 = 0to see how these principles come together.

Discriminant

The discriminant is a crucial component of the quadratic formula, acting as a determinant of the nature and number of the roots for the equation. It is given by the expression [...theres nothing think better formula here than d = b^2 - 4ac,]
The value of the discriminant can reveal different properties of the roots:

• If the discriminant d > 0, there are two distinct real roots.
• If d = 0, there is one real root (or a repeated root).
• If d < 0, the roots are complex or imaginary.

In our example, the discriminant is calculated as: =(-6)^2 - 4(2)(3) = 36 - 24 = 12. Since [d] is positive, we know there are two distinct real roots.

Roots of Quadratic Equations

The roots (or solutions) of a quadratic equation are the values of the variable that satisfy the equation. These can be found by solving the quadratic formula:
\[a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In our example, the quadratic formula application yields:
\[a = \frac{-(-6) \pm 2\sqrt{3}}{2(2)}\]
Upon simplification:
\[a = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3}{2} \pm \frac{\sqrt{3}}{2}\]
This unravels two specific roots:

• [quadraticmath3] \
• [quadraticmath3]
• These are the two values of a where the original equation holds true..
  

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Algebraic Solutions

Algebraic solutions refer to solving equations using algebraic manipulations. For quadratic equations, the algebraic solution typically involves using the quadratic formula. Here is the step-by-step algebraic solution for our quadratic equation, 2a^2 - 6a + 3 = 0:

1. Identify the coefficients: a = 2, b = -6, c = 3.
2. Calculate the discriminant: (b^2 - 4ac) = 12.
3. Compute the square root of the discriminant: \[\sqrt{12} = 2\sqrt{3}\]
4. Substitute into the quadratic formula: \[a = \frac{6 \pm 2\sqrt{3}}{4}\]


5. Simplify to find the roots: \[a = \frac{3}{2} \pm \frac{\sqrt{3}}{2}\]
These algebraic steps systematically break down the problem to reach the solution, showing that anyone can master quadratic equations with practice and patience.