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Chapter 5: Problem 68

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

Short Answer

Expert verified
The solutions to the quadratic equation are \( u = \frac{1+\sqrt{5}}{2} \) and \( u = \frac{1-\sqrt{5}}{2} \).

Step by step solution

01

Identify Variables

Identify variables a, b, and c from the equation. Here a=1, b=-1 and c=-1.
02

Substitute Variables into Quadratic Formula

Substitute a, b, and c into the quadratic formula: \( u = \frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)} \).
03

Simplify under the Square Root

Simplify the expression under the square root to obtain: \( u = \frac{1\pm\sqrt{1+4}}{2} \), which simplifies further to \( u = \frac{1\pm\sqrt{5}}{2} \).
04

Calculate the Solutions

Now we have the two solutions: \( u = \frac{1+\sqrt{5}}{2} \) and \( u = \frac{1-\sqrt{5}}{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Formula
The quadratic formula is a powerful tool used to solve quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). This formula gives you a straightforward way to find the solutions for \( x \).
The formula is:
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
In this context:
  • \( a \) is the coefficient of \( x^2 \)
  • \( b \) is the coefficient of \( x \)
  • \( c \) is the constant term
Using the formula involves identifying these coefficients and substituting them into the equation. The "\( \pm \)" symbol indicates there are generally two solutions to consider.
Discriminant
The discriminant is a key part of the quadratic formula. It is the expression under the square root: \( b^2 - 4ac \). This component determines the nature of the solutions to a quadratic equation. Here's why the discriminant is important:
  • If the discriminant is positive, \( b^2 - 4ac > 0 \), the equation has two distinct real solutions.
  • If the discriminant is zero, \( b^2 - 4ac = 0 \), the equation has exactly one real solution (a repeated root).
  • If the discriminant is negative, \( b^2 - 4ac < 0 \), the equation has no real solutions, but two complex ones.
In the given problem, \( b^2 - 4ac \) equals 5, which is positive, indicating that there are two real solutions.
Solutions of Equations
The result of using the quadratic formula are the solutions to the equation. In solving these, we substitute the values of \( a \), \( b \), and \( c \) into the quadratic formula and reduce the expression.For example, in our exercise we get:
  • \( u = \frac{1 + \sqrt{5}}{2} \)
  • \( u = \frac{1 - \sqrt{5}}{2} \)
These are the solutions for \( u^2 - u - 1 = 0 \). Each solution (root) represents a value that satisfies the original quadratic equation. Remember, the solutions may be real or complex, and using the discriminant helps us predict the nature of these solutions before actually calculating them.

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Most popular questions from this chapter

In Exercises \(147-151,\) use a graphing utility to approximate the solutions of each equation in the interval \([0,2 \pi) .\) Round to the nearest hundredth of a radian. $$\cos x=x$$

In Exercises \(147-151,\) use a graphing utility to approximate the solutions of each equation in the interval \([0,2 \pi) .\) Round to the nearest hundredth of a radian. $$\sin 2 x=2-x^{2}$$

Exercises \(166-168\) will help you prepare for the material covered in the first section of the next chapter. Solve each equation by using the cross- products principle to clear fractions from the proportion: $$ \text { If } \frac{a}{b}=\frac{c}{d}, \text { then } a d=b c .(b \neq 0 \text { and } d \neq 0) $$ Round to the nearest tenth. $$\text { Solve for } B: \frac{51}{\sin 75^{\circ}}=\frac{71}{\sin B}$$

In Exercises \(85-96,\) use a calculator to solve each equation, correct to four decimal places, on the interval \([0,2 \pi)\) $$3 \cos ^{2} x-8 \cos x-3=0$$

Determine whether each statement makes sense or does not make sense, and explain your reasoning. I used a half-angle formula to find the exact value of \(\cos 100^{\circ}\)
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