Problem 71 - Algebra 1: Concepts and Skills

Question

Use the quadratic formula to solve the equation. If the solution involves radicals, round to the nearest hundredth. $$x^{2}-6 x+1=0$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 5.83\) and \(x = 0.17\).

Step 1: Identification of the Coefficients

From the given quadratic equation \(x^{2}-6x+1=0\), identify the coefficients: \(a=1\), \(b=-6\), and \(c=1\)

Step 2: Substitue Coefficients Into Quadratic Formula

Substitute the coefficients into the quadratic formula. This yields \(x= \frac{-(-6) \pm \sqrt{(-6)^{2}-4*1*1}}{2*1}\)

Step 3: Simplify Under The Radical

Simplify the calculation under the square root, also known as the discriminant \((b^{2}-4ac)\). This simplification gives \(x=\frac{6 \pm \sqrt{36-4}}{2}\)

Step 4: Calculate the final solutions

Finally, we calculate the values for 'x'. This calculation gives \(x=\frac{6 \pm \sqrt{32}}{2}\), which simplifies to \(x=3 \pm \sqrt{8}\). To approximate the solutions up to the nearest hundredth, it will be \(x = 3\pm 2.83\). Therefore, the solutions will be \(x = 5.83\) and \(x=0.17\).

Key Concepts

Quadratic Equation

A quadratic equation is an algebraic expression that can be represented as \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, while \(x\) is the variable we want to solve. It is called "quadratic" because "quad" typically refers to four, and these polynomials are of the second degree, involving \(x^2\) as the highest power. In our given equation, \(x^2 - 6x + 1 = 0\), the values are:

• \(a = 1\)
• \(b = -6\)
• \(c = 1\)

The primary aim is to find the value of \(x\) that satisfies this equation. Quadratic equations can often have two solutions due to the square term. This is where the quadratic formula becomes a vital tool for finding these solutions.

Discriminant

The discriminant is a component of the quadratic formula that determines the nature of the roots of the equation. It is given by the formula: \(b^2 - 4ac\). The value obtained from this calculation gives insight into the solutions of the quadratic equation:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root (or a repeated root).
• If the discriminant is negative, the roots are complex or imaginary numbers.

In our example, we found the discriminant by simplifying \((-6)^2 - 4 \times 1 \times 1\), which resulted in \(36 - 4 = 32\). Since 32 is positive, our quadratic equation has two real solutions, which we can proceed to find using the quadratic formula.

Radicals

Radicals involve square roots and are often encountered when solving quadratic equations using the quadratic formula. In the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), the expression under the square root, \(\sqrt{b^2 - 4ac}\), is what introduces radicals.
To solve the equation \(x^2 - 6x + 1 = 0\) using the quadratic formula, we simplified the expression under the radical to \(\sqrt{32}\). The square root of 32 is not a perfect square, which means the solutions will involve radicals. These solutions can then be approximated using decimals for practical purposes.
For our equation, simplifying led to \(3 \pm \sqrt{8}\), which approximates to \(3 \pm 2.83\). Thus, the rounded solutions are \(x = 5.83\) and \(x = 0.17\). Radicals allow us to express these non-perfect square roots easily in terms of numbers that are more interpretable.