Question

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Solve \(3 x^{2}+x=1\) by using the quadratic formula.

Short answer

\(x = \frac{-1 + \sqrt{13}}{6}\) and \(x = \frac{-1 - \sqrt{13}}{6}\)

Step-by-step solution

Identify coefficients

For the quadratic equation in standard form, ax^2 + bx + c = 0,identify the coefficients. In the equation 3x^2 + x = 1, first rearrange it to: 3x^2 + x - 1 = 0.Therefore, the coefficients are: a = 3,b = 1,c = -1.

Quadratic formula

Write down the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].

Calculate the discriminant

Calculate the discriminant (the value inside the square root) using the coefficients identified in Step 1: \[b^2 - 4ac = 1^2 - 4(3)(-1) = 1 + 12 = 13\].

Apply the quadratic formula

Substitute the coefficients and the discriminant into the quadratic formula to solve for x: \[x = \frac{-1 \pm \sqrt{13}}{6}\].

Simplify the solutions

This results in two possible solutions for x: \[x = \frac{-1 + \sqrt{13}}{6}\] and \[x = \frac{-1 - \sqrt{13}}{6}\].

Key concepts

quadratic equations

A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). Quadratic equations are important because they model a wide range of real-world problems, from physics to finance.

Key characteristics of quadratic equations include:

• The highest degree of the variable \(x\) is 2.
• The graph of a quadratic equation is a parabola which opens upwards if \(a > 0\) and downwards if \(a < 0\).
• It can have zero, one, or two real solutions depending on the value of the discriminant (more on that later).

In terms of our exercise, the given equation was \(3x^2 + x = 1\). Notice that to fit it into the standard form, we rearranged it to \(3x^2 + x - 1 = 0\), making it easier to identify the coefficients.

solving quadratic equations

Solving quadratic equations can be done in several ways, but using the quadratic formula is one of the most reliable methods. The quadratic formula is given by:
\ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \
Here’s a quick breakdown of how it's used:

• Identify the coefficients \(a\), \(b\), and \(c\) from the quadratic equation in standard form \(ax^2 + bx + c = 0\).
• Calculate the discriminant \(b^2 - 4ac\).
• Substitute the values of \(a\), \(b\), and the discriminant into the formula.
• Solve for \(x\) by simplifying the values inside the square root and the fraction.

In our exercise, we had \(a=3\), \(b=1\), and \(c=-1\). Plugging these into the quadratic formula, we got: \ x = \frac{-1 \pm \sqrt{13}}{6} \
This resulted in two solutions: \ x = \frac{-1 + \sqrt{13}}{6} \ and \ x = \frac{-1 - \sqrt{13}}{6}.

discriminant

The discriminant is a key part of the quadratic formula, represented by \(b^2 - 4ac\). It tells us the nature of the roots of the quadratic equation:

• If the discriminant is positive ( \(b^2 - 4ac > 0\) ), there are two distinct real roots.
• If the discriminant is zero ( \(b^2 - 4ac = 0\) ), there is exactly one real root (a repeated root).
• If the discriminant is negative ( \(b^2 - 4ac < 0\) ), there are no real roots, but two complex roots.

In our problem, the discriminant was calculated as: \ 1^2 - 4(3)(-1) = 1 + 12 = 13 \
Since \(13 > 0\), we concluded that there are two distinct real roots for our equation. This discriminant helps students understand whether they need to consider real roots or look at complex numbers based on the nature of the problem they are solving.