Problem 66 - Precalculus Essentials

Question

Will help you prepare for the material covered in the next section. $$\text { Solve: } 2\left(1-u^{2}\right)+3 u=0$$

Step-by-Step Solution

Answer

The quadratic equation \(2u^2 - 3u + 2 = 0\) has no real solution.

Step 1: Rewrite the equation in standard form

Rearrange the terms to express the equation as a standard quadratic equation. This involves distributing the 2 in the brackets and moving all terms to one side. So, the equation becomes \(2-2u^2+3u=0\) or rewriting it in a standard format \(2u^2 - 3u + 2 = 0\).

Step 2: Using the Quadratic Formula

Plug the coefficients of \(u^2\), \(u\), and the constant into the Quadratic Formula \(-b \pm \sqrt{b^2-4ac} \over 2a\), where \(a=2\), \(b=-3\), and \(c=2\). So do: \(u=\frac{-(-3)\pm\sqrt{(-3)^2-4*2*2}}{2*2}\).

Step 3: Simplify the equation

Perform the operations under the square root, and simplify to obtain the roots. Simplifying gives: \(u=\frac{3\pm\sqrt{9-16}}{4} = \frac{3\pm\sqrt{-7}}{4}\). Since we're dealing with real numbers, the square root of a negative number doesn't exist. Thus, the equation has no real solution.

Key Concepts

Standard Form of a Quadratic Equation

To fully understand the steps of solving a quadratic equation, let's start by discussing the 'standard form'. The standard form of a quadratic equation is written as \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). The term \(x\) represents the unknown variable we aim to solve for. Writing an equation in standard form is a crucial first step because it provides a clear structure that makes it easier to apply solution strategies, such as factoring or using the quadratic formula.
In our exercise, the given equation \(2\left(1-u^2\right)+3u=0\) needed rearrangement. By distributing and rearranging the expression into \(2u^2 - 3u + 2 = 0\), we put it in standard form. This sets the stage for solving the equation using the quadratic formula.

Quadratic Formula

The quadratic formula is a critical tool for solving quadratic equations in standard form \(ax^2 + bx + c = 0\). It allows us to find the values of \(x\) by using the coefficients \(a\), \(b\), and \(c\). The formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

• The term \(-b\) starts the equation.
• The \(\pm\) symbol indicates two potential solutions due to the square root.
• The expression within the square root, \(b^2 - 4ac\), is known as the discriminant.

In the exercise, after substituting \(a = 2\), \(b = -3\), and \(c = 2\) into the formula, we calculate \(x = \frac{3 \pm \sqrt{-7}}{4}\). Notably, the discriminant \(b^2 - 4ac\) resulted in a negative value, revealing that the equation does not have real solutions.

Real Numbers and the Nature of Solutions

Real numbers include all the numbers on the number line like whole numbers, fractions, and decimals. Importantly, they do not include complex or imaginary numbers, such as the square root of negative numbers.
In quadratic equations, the type of solutions we find depends on the discriminant \(b^2 - 4ac\):

• If positive, there are two distinct real solutions.
• If zero, there is exactly one real solution.
• If negative, as with our exercise, there are no real solutions.

So in the exercise, \(b^2 - 4ac = -7\), indicating no real roots. This means the solutions involve imaginary numbers, which aren't considered real. Hence, our process shows that no real solutions exist for the given equation.