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Chapter 1: Problem 66

Solve each equation in the complex number system. $$ x^{2}-2 x+5=0 $$

Short Answer

Expert verified
The solutions are \(x = 1 + 2i\) and \(x = 1 - 2i\).

Step by step solution

01

- Identify the quadratic equation

The given quadratic equation is \(x^{2} - 2x + 5 = 0\). This is in the standard form of a quadratic equation \(ax^2 + bx + c = 0\) where \(a = 1\), \(b = -2\), and \(c = 5\).
02

- Apply the quadratic formula

Use the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Here, \(a = 1\), \(b = -2\), and \(c = 5\). Substitute these values into the formula.
03

- Calculate the discriminant

Calculate the discriminant \(\Delta = b^2 - 4ac\). Here, \(b = -2\), \(a = 1\), and \(c = 5\). So, \(\Delta = (-2)^2 - 4(1)(5) = 4 - 20 = -16\).
04

- Solve for x using the quadratic formula

Since the discriminant is negative, the solutions will be complex numbers. Substitute \(b = -2\), \(a = 1\), and the discriminant \(\Delta = -16\) into the quadratic formula: \[ x = \frac{-(-2) \pm \sqrt{-16}}{2(1)} \]This simplifies to \[ x = \frac{2 \pm 4i}{2} \]which further simplifies to \[ x = 1 \pm 2i \].

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Key Concepts

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

quadratic formula
To solve any quadratic equation in the form of \(ax^2 + bx + c = 0\), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
This formula allows us to find the solutions for \(x\) by substituting the coefficients \(a\), \(b\), and \(c\) from the quadratic equation.
In our exercise, the equation is \(x^2 - 2x + 5 = 0\). Here, \(a = 1\), \(b = -2\), and \(c = 5\). Substituting these values into the formula we get:
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(5)}}{2(1)} \].
This step is crucial in solving any quadratic equation and understanding it well.
complex numbers
When solving quadratic equations, sometimes we encounter negative discriminants, resulting in complex solutions.
A complex number consists of a real part and an imaginary part and is written as \(a + bi\), where \(i\) is the imaginary unit defined by \(i^2 = -1\).
In our exercise, the discriminant \(\Delta\) was \(-16\), which means we have to involve the imaginary unit.
By substituting into the quadratic formula, we get:
\[ x = \frac{2 \pm 4i}{2} \]
which simplifies to \(x = 1 \pm 2i\), these are our complex solutions.
Understanding complex numbers is key for many advanced math topics.
discriminant
The discriminant \(\Delta\) in the quadratic formula, \(\Delta = b^2 - 4ac\), determines the nature of the roots.
- If \(\Delta > 0\), there are two distinct real roots.
- If \(\Delta = 0\), there is exactly one real root.
- If \(\Delta < 0\), there are two complex roots.
In our exercise, with \(a = 1\), \(b = -2\), and \(c = 5\), we calculated \(\Delta\) to be \(\Delta = -16\). Because \(\Delta\) is negative, it indicates that the quadratic equation has two complex roots.
This is essential to understand because it helps predict the type of solutions without fully solving the equation first.

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Most popular questions from this chapter

Cyclists Two cyclists leave a city at the same time, one going east and the other going west. The westbound cyclist bikes 5 mph faster than the eastbound cyclist. After 6 hours they are 246 miles apart. How fast is each cyclist riding?

The period of a pendulum is the time it takes the pendulum to make one full swing back and forth. The period \(T,\) in seconds, is given by the formula \(T=2 \pi \sqrt{\frac{l}{32}}\) where \(l\) is the length, in feet, of the pendulum. In 1851 , Jean Bernard Leon Foucault win Paris. The period of Foucault's pendulum was approximately 16.5 seconds. What was its length?

The distance to the surface of the water in a well can sometimes be found by dropping an object into the well and measuring the time elapsed until a sound is heard. If \(t_{1}\) is the time (measured in seconds) that it takes for the object to strike the water, then \(t_{1}\) will obey the equation \(s=16 t_{1}^{2}\), where \(s\) is the distance (measured in feet). It follows that \(t_{1}=\frac{\sqrt{s}}{4}\). Suppose that \(t_{2}\) is the time that it takes for the sound of the impact to reach your ears. Because sound waves are known to travel at a speed of approximately 1100 feet per second, the time \(t_{2}\) to travel the distance \(s\) will be \(t_{2}=\frac{s}{1100} .\) See the illustration. Now \(t_{1}+t_{2}\) is the total time that elapses from the moment that the object is dropped to the moment that a sound is heard. We have the equation $$ \text { Total time elapsed }=\frac{\sqrt{s}}{4}+\frac{s}{1100} $$ Find the distance to the water's surface if the total time elapsed from dropping a rock to hearing it hit water is 4 seconds.

Find the real solutions, if any, of each equation. $$ 3 x^{2}+7 x-20=0 $$

Find the real solutions, if any, of each equation. Use any method. $$ 10 x^{2}-19 x-15=0 $$
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