/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Give all solutions of each nonli... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Give all solutions of each nonlinear system of equations, including those with nonreal complex components. $$\begin{aligned} &5 x^{2}-y^{2}=0\\\ &3 x^{2}+4 y^{2}=0 \end{aligned}$$

Short Answer

Expert verified
The solution is (0, 0).

Step by step solution

01

Analyze the given system of equations

The given system of equations are: 1) \(5x^{2} - y^{2} = 0\) 2) \(3x^{2} + 4y^{2} = 0\). We need to find solutions for \(x\) and \(y\) that satisfy both equations.
02

Solve the first equation for y

From the first equation, solve for \(y^{2}\): \(5x^{2} - y^{2} = 0\) This simplifies to: \(y^{2} = 5x^{2}\).
03

Substitute y^2 into the second equation

Substitute \(y^{2} = 5x^{2}\) into the second equation: \(3x^{2} + 4(5x^{2}) = 0\). This simplifies to: \(3x^{2} + 20x^{2} = 0\), which further simplifies to: \(23x^{2} = 0\).
04

Solve for x

From the equation \(23x^{2} = 0\), solve for \(x\): \(x^{2} = 0\) So, \(x = 0\).
05

Solve for y using x = 0

With \(x = 0\), substitute back into the equation \(y^{2} = 5x^{2}\): \(y^{2} = 5(0)^{2}\) Therefore, \(y = 0\).
06

Check for complex solutions

Considering complex solutions: If \(x\) and \(y\) are nonreal (complex), \(x = ai\) (where \(i\) is the imaginary unit) leads to nonreal solutions. Re-evaluate using \(x = ai\): 1) \(5(ai)^{2} - y^{2} = 0\): \(-5a^{2} - y^{2} = 0\) This would imply \(y = bi\) (nonreal solution does not need re-evaluation in this equation set). Thus the valid real solution remains \(x = 0, y = 0\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Complex Solutions
Complex solutions arise when variables in an equation take nonreal values. In the context of a nonlinear system, this means solutions such as imaginary numbers. When working with complex solutions, it's crucial to understand imaginary units and how they behave in equations.

In our exercise, solving the equations led to a set of real solutions first, with both variables equating to zero. We explored potential complex solutions by assuming the variable could be complex.

Considering the variable as a complex number, say, in the form of \(x = ai\), and substituting back into the equations reveals nonreal components.

This approach might make complex solutions visible, showing that complex roots come into play when real solutions are insufficient or nonexistent.

Remember, the key steps involve assuming nonreal parts for variables, typically using the imaginary unit for substitution. You should then re-solve the equations to confirm if valid nonreal solutions exist.
Solving Quadratic Equations
In many problems involving nonlinear systems, you often need to solve quadratic equations. A quadratic equation takes the general form \(ax^2 + bx + c = 0\). To find the roots, you can use the quadratic formula:

    \t
  • \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
The discriminant, \(b^2 - 4ac\), determines the nature of the roots:
    \t
  • If the discriminant is positive, you get two distinct real roots.
  • \t
  • If it's zero, there's one real root.
  • \t
  • If negative, you find complex roots.
In our nonlinear system, we started by isolating one variable and substituting it into the second equation, leading us to solve equivalent quadratic forms.

In many cases, you need to recognize when to apply quadratic formulas versus simplifying directly for roots. Here, for example, getting \(y = \pm \sqrt{5x^2}\) made further substitution straightforward.

Solving these using substitution allows you to handle both the real and complex solutions efficiently, as we saw by reassessing the roots under both real and complex assumptions.
Imaginary Unit
The imaginary unit, denoted as \(i\), is a fundamental concept in handling complex numbers. Defined as \(i = \sqrt{-1}\), it makes working with square roots of negative numbers possible.

When you square \(i\), you get:
\[ i^2 = -1 \
\]This property turns the imaginary unit into a powerful tool for solving equations that don't have real solutions. Complex numbers are written in the form \(a + bi\), where \(a\) and \(b\) are real numbers.

In our exercise, when analyzing potential complex solutions for the system \(5x^{2} - y^{2} = 0\) and \(3x^{2} + 4y^{2} = 0\), introducing \(x = ai\) allowed us to handle the system with nonreal components effectively.

For a practical approach, recognize the need to incorporate imaginary units as soon as the equation suggests the impossibility of real roots. In essence, the imaginary unit helps expand the solution space, making complex numbers approachable and manageable.

Understanding \(i\) helps bridge gaps in solutions that could otherwise be unsolvable under real-number constraints.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Storage Capacity An office manager wants to buy some filing cabinets. He knows that cabinet A costs \(\$ 10\) each, requires \(6 \mathrm{ft}^{2}\) of floor space, and holds \(8 \mathrm{ft}^{3}\) of files. Cabinet B costs \(\$ 20\) each, requires \(8 \mathrm{ft}^{2}\) of floor space, and holds \(12 \mathrm{ft}^{3}\) of files. He can spend no more than \(\$ 140\) due to budget limitations, and his office has room for no more than \(72 \mathrm{ft}^{2}\) of cabinets. He wants to maximize storage capacity within the limits imposed by funds and space. How many of each type of cabinet should he buy?

Use the shading capabilities of your graphing calculator to graph each inequality or system of inequalities. $$\begin{aligned}&y \geq|x+2|\\\&y \leq 6\end{aligned}$$

Solve each system by using the inverse of the coefficient matrix. $$\begin{aligned} &x+3 y=-12\\\ &2 x-y=11 \end{aligned}$$

Gasoline Revenues The manufacturing process requires that oil refineries manufacture at least 2 gal of gasoline for each gallon of fuel oil. To meet the winter demand for fuel oil, at least 3 million gal per day must be produced. The demand for gasoline is no more than 6.4 million gal per day. If the price of gasoline is \(\$ 2.90\) per gal and the price of fuel oil is \(\$ 2.50\) per gal, how much of each should be produced to maximize revenue?

Solve each problem. Tire Sales The number of automobile tire sales is dependent on several variables. In one study the relationship among annual tire sales \(S\) (in thousands of dollars), automobile registrations \(R\) (in millions), and personal disposable income \(I\) (in millions of dollars) was investigated. The results for three years are given in the table. To describe the relationship among these variables, we can use the equation $$ S=a+b R+c l $$ where the coefficients \(a, b,\) and \(c\) are constants that must be determined before the equation can be used. (Source: Jarrett, J., Business Forecasting Methods, Basil Blackwell, Ltd.) (a) Substitute the values for \(S, R,\) and \(I\) for each year from the table into the equation \(S=a+b R+c I,\) and obtain three linear equations involving \(a, b,\) and \(c\) (b) Use a graphing calculator to solve this linear system for \(a, b,\) and \(c .\) Use matrix inverse methods. (c) Write the equation for \(S\) using these values for the coefficients. (d) If \(R=117.6\) and \(I=310.73,\) predict \(S .\) (The actual value for \(S\) was \(11,314 .\) ) (e) If \(R=143.8 \text { and } I=829.06, \text { predict } S . \text { (The actual value for } S \text { was } 18,481 .)\) $$\begin{array}{|c|c|c|} \hline S & R & I \\ \hline 10,170 & 112.9 & 307.5 \\\ \hline 15,305 & 132.9 & 621.63 \\ \hline 21,289 & 155.2 & 1937.13 \\\ \hline \end{array}$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.