Problem 20 - Precalculus: Functions and Graphs

Question

Solve by using the quadratic formula. \(x^{2}-6 x-3=0\)

Step-by-Step Solution

Answer

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

Step 1: Identify Coefficients

Identify the coefficients of the quadratic equation. In the equation \(x^2 - 6x - 3 = 0\), the coefficients are \(a = 1\), \(b = -6\), and \(c = -3\).

Step 2: Write the Quadratic Formula

The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] where \(a\), \(b\), and \(c\) are coefficients from the quadratic equation.

Step 3: Substitute Coefficients into Formula

Substitute \(a = 1\), \(b = -6\), and \(c = -3\) into the quadratic formula:\[x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1}\]

Step 4: Simplify the Expression

Simplify the expression inside the square root and the rest of the quadratic formula:- Compute \((-6)^2 = 36\)- Compute \(4 \cdot 1 \cdot (-3) = -12\)- Calculate the discriminant: \(36 + 12 = 48\)- Hence, the formula becomes:\[x = \frac{6 \pm \sqrt{48}}{2}\]

Step 5: Simplify the Square Root

Simplify \(\sqrt{48}\):- \(\sqrt{48} = \sqrt{16 \cdot 3} = \sqrt{16} \cdot \sqrt{3} = 4\sqrt{3}\)- Substitute back into the equation:\[x = \frac{6 \pm 4\sqrt{3}}{2}\]

Step 6: Final Simplification

Simplify the fractions:- Divide the whole expression by 2:\[x = \frac{6}{2} \pm \frac{4\sqrt{3}}{2}\]- This simplifies to:\[x = 3 \pm 2\sqrt{3}\]

Key Concepts

coefficients of a quadratic equation

In a quadratic equation, the coefficients are the numbers that multiply the variables in the expression. A typical quadratic equation is written in the form:

• \(ax^2 + bx + c = 0\)

The letters \(a\), \(b\), and \(c\) represent the coefficients.

• \(a\) is the coefficient of \(x^2\), which is the quadratic term.
• \(b\) is the coefficient of \(x\), which is the linear term.
• \(c\) is the constant term, or the term without an \(x\).

It's important that \(a eq 0\), because if \(a\) were zero, the equation would no longer be quadratic. In the equation given in the exercise "\(x^2 - 6x - 3 = 0\)", identify the following:

• The coefficient \(a = 1\) because it's understood there's a 1 in front of \(x^2\).
• The coefficient \(b = -6\).
• The constant term \(c = -3\).

Recognizing these coefficients is crucial as they are used throughout the process of applying the quadratic formula.

discriminant calculation

The discriminant is an important part of the quadratic formula. It is the expression under the square root sign. For a quadratic equation represented by:
\(ax^2 + bx + c = 0\)
The discriminant \(D\) is calculated as:

• \(b^2 - 4ac\)

This value will tell us about the nature of the roots of the quadratic equation.

• If \(D > 0\), the quadratic equation has two distinct real roots.
• If \(D = 0\), it has exactly one real root (also called a repeated or double root).
• If \(D < 0\), the roots are complex or imaginary.

In the exercise "\(x^2 - 6x - 3 = 0\)", we substitute the coefficients into the discriminant calculation:
\[b^2 - 4ac = (-6)^2 - 4 \cdot 1 \cdot (-3) = 36 + 12 = 48\]
Since our discriminant here is positive (\(48\)), the quadratic equation has two distinct real roots, which we can then find using the quadratic formula.

simplifying square roots

Simplifying square roots is an important skill needed when solving quadratic equations, especially when the discriminant is not a perfect square. When faced with a number like 48 inside a square root, we reduce it to its simplest form by factoring it into a product of squares. To simplify \(\sqrt{48}\):

• Factor 48 into its prime factors: \(48 = 16 \times 3\)
• Recognize that \(16\) is a perfect square, as \(16 = 4^2\).
• Write \(\sqrt{48}\) as \(\sqrt{16 \times 3}\).
• Simplify to \(\sqrt{16} \times \sqrt{3}\) which leads to \(4\sqrt{3}\).

This results in \(\sqrt{48} = 4\sqrt{3}\), an accurate and simplified expression. Ensuring that a square root is fully simplified is crucial to accurately finding the roots of a quadratic equation. In the exercise's final steps, this leads to the expression \(x = 3 \pm 2\sqrt{3}\). Clarity in simplifying square roots is essential for proceeding to the last step—solving the equation completely.