Problem 39 - Precalculus

Question

Use the quadratic formula to solve each equation. In Exercises \(34-39,\) give two forms for each solution: an expression containing a radical and a calculator approximation rounded off to two decimal places. $$\sqrt{3} x^{2}+\sqrt{3}=6 x$$

Step-by-Step Solution

Answer

The solutions are \(x = \sqrt{3} \pm \sqrt{2}\), approximately \(x \approx 2.73\) and \(x \approx 0.32\).

Step 1: Write the equation in standard form

The given quadratic equation is \(\sqrt{3}x^2 + \sqrt{3} = 6x\). To use the quadratic formula, we need to rearrange this into its standard form, \(ax^2 + bx + c = 0\). Subtract \(6x\) from both sides: \(\sqrt{3}x^2 - 6x + \sqrt{3} = 0\).

Step 2: Identify coefficients

For the quadratic equation \(ax^2 + bx + c = 0\), identify the coefficients: \(a = \sqrt{3}\), \(b = -6\), and \(c = \sqrt{3}\).

Step 3: Apply the quadratic formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the values of \(a\), \(b\), and \(c\): \(x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(\sqrt{3})(\sqrt{3})}}{2(\sqrt{3})}\).

Step 4: Simplify the formula expression

Calculate the discriminant: \((-6)^2 - 4(\sqrt{3})(\sqrt{3}) = 36 - 12 = 24\). The expression becomes \(x = \frac{6 \pm \sqrt{24}}{2\sqrt{3}}\).

Step 5: Simplify the square root

Simplify \(\sqrt{24}\): \(\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}\). The expression is now \(x = \frac{6 \pm 2\sqrt{6}}{2\sqrt{3}}\).

Step 6: Rationalize and simplify the solution

Divide both terms in the numerator by 2: \(x = \frac{3 \pm \sqrt{6}}{\sqrt{3}}\). Multiply numerator and denominator by \(\sqrt{3}\) to rationalize: \(x = \frac{3\sqrt{3} \pm \sqrt{18}}{3}\). Simplify \(\sqrt{18}\) as \(3\sqrt{2}\), so the solution is \(x = \frac{3\sqrt{3} \pm 3\sqrt{2}}{3}\), which simplifies to \(x = \sqrt{3} \pm \sqrt{2}\).

Step 7: Calculate decimal approximation

Use a calculator to approximate \(x = \sqrt{3} + \sqrt{2}\) and \(x = \sqrt{3} - \sqrt{2}\). Approximations are \(x \approx 2.73\) and \(x \approx 0.32\) respectively.

Key Concepts

Quadratic Equations

Quadratic Equations are a fundamental part of algebra, and they frequently appear in both academic exercises and real-life applications. These equations take the general form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown variable we aim to solve for.

The term "quadratic" comes from "quad," meaning square, because the variable \( x \) is squared. This type of equation can have two solutions, one solution, or sometimes no real solution, depending on the nature of its coefficients and the equation's discriminant.

To effectively tackle quadratic equations, understanding their behavior and potential solutions is invaluable. In many cases, the quadratic formula provides the fastest solution path.

Standard Form

Before applying techniques like the quadratic formula, converting a quadratic equation into its standard form is crucial. Standard form is expressed as \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are coefficients.

In our exercise, the initial equation is \( \sqrt{3}x^2 + \sqrt{3} = 6x \). By moving all terms to one side and setting the equation to zero, it becomes \( \sqrt{3}x^2 - 6x + \sqrt{3} = 0 \).

Rearranging into this form helps in identifying the coefficients necessary for solving the equation using the quadratic formula. This setup is particularly helpful because it simplifies the process and highlights the roles of each component in the quadratic.

Discriminant

The discriminant is a powerful tool that tells us about the nature of the roots of a quadratic equation without having to find them. It is part of the quadratic formula within the square root: \( b^2 - 4ac \).

Depending on the value of the discriminant:

• If it's positive, the quadratic equation will have two distinct real roots.
• If it's zero, there will be exactly one real root, also known as a repeated root.
• If it's negative, there are no real roots, and the solutions are complex.

In this exercise, substituting the values onto the discriminant formula gives \( 36 - 12 = 24 \), which is positive. Thus, the equation has two distinct real roots, aligning perfectly with the two forms we were asked to find.

Simplification

Simplification is about reducing the complexity of mathematical expressions. This makes them easier to interpret, solve, or approximate. In our exercise, simplification involves several steps.

First, the quadratic formula yields the expression \( x = \frac{6 \pm \sqrt{24}}{2\sqrt{3}} \). The next move is simplifying \( \sqrt{24} \) by expressing it as \( 2\sqrt{6} \), leading to \( x = \frac{6 \pm 2\sqrt{6}}{2\sqrt{3}} \).
Continuing, reduce by dividing, \( x = \frac{3 \pm \sqrt{6}}{\sqrt{3}} \). Finally, rationalizing the denominator gives \( x = \sqrt{3} \pm \sqrt{2} \).

Simplification is crucial for making expressions more comprehensible and manageable especially when you're solving math problems manually or using them in further applications.

Calculator Approximation

In mathematical problems, especially those requiring practical applications or approximations, a calculator can be a very handy tool to yield numerical solutions.

With the simplified radical expressions of our solution as \( x = \sqrt{3} + \sqrt{2} \) and \( x = \sqrt{3} - \sqrt{2} \), we can approximate these using a calculator. The results are approximately 2.73 and 0.32, respectively.

These approximations provide a tangible sense of the values involved, which is often required in practical fields. Calculators ensure precision by accurately computing square roots and other mathematical functions, key when real-world precision is necessary beyond symbolic solutions.