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Chapter 8: Problem 12

In Exercises \(1-34\), find all real solutions of the system of equations. If no real solution exists, so state. $$\left\\{\begin{array}{c} 4 x^{2}+y=2 \\ -4 x+y=3 \end{array}\right.$$

Short Answer

Expert verified
The real solution to the system of equations is (-0.5, 1)

Step by step solution

01

Write down the system of equations

The system of equations is given as: \(4x^2 + y = 2\) and \(-4x + y = 3\)
02

Express y from one equation

Choose the simpler equation to express y, \(-4x + y = 3\), thus, \(y = 3 + 4x\)
03

Substitute y in the other equation

Substitute \(y = 3 + 4x\) into \(4x^2 + y = 2\) which gives: \(4x^2 + 3 + 4x = 2\)
04

Simplify the equation

Simplify the equation above to look like a standard quadratic equation as \(4x^2 + 4x + 3 - 2 = 0\), which finally gives: \(4x^2 + 4x + 1 = 0\)
05

Solve the quadratic equation

The quadratic equation \(4x^2 + 4x + 1 = 0\) can be written as \(2x + 1\)^2 = 0. Solving it we find that \(x = -0.5\)
06

Substitute x into the other equation

Substitute \(x = -0.5\) into \(y = 3 + 4x\) which gives: \(y = 3 + 4(-0.5)\), thus, \(y = 1\)

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Key Concepts

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

Quadratic Equations
Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). These equations can depict many real-world phenomena, such as projectile motion and areas.
  • The highest power of the variable \( x \) is 2, making it a "quadratic," which comes from the Latin word "quadratus," meaning "square."
  • Solving a quadratic equation usually involves finding the value(s) of \( x \) that make the equation true. These values are called "roots" or "solutions," and there can be up to two distinct solutions.
  • To solve a quadratic equation, you can use several methods, such as factoring, completing the square, or using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Each method requires different conditions and is suited to different types of equations.
In our case, we transformed the equation \( 4x^2 + 4x + 1 = 0 \) into \( (2x + 1)^2 = 0 \), suggesting a perfect square, leading directly to its root \( x = -0.5 \).
Substitution Method
The substitution method is a technique for solving systems of equations, particularly useful when dealing with linear equations and systems involving quadratic equations.
  • This method involves solving one of the equations for one variable and then substituting this expression into the other equation. This reduces the system into a single equation with one variable, making it easier to solve.
  • In the given system, we first solved for \( y \) in the linear equation: \(-4x + y = 3\), rewriting it as \( y = 3 + 4x \).
  • Then, this expression for \( y \) was substituted into the quadratic equation \( 4x^2 + y = 2 \), allowing us to focus on solving just one equation: \( 4x^2 + 4x + 1 = 0 \).
The substitution method is powerful because it consolidates information, focusing our efforts on solving a simpler problem.
Real Solutions
When solving equations, a "real solution" refers to a solution that is a real number, as opposed to an imaginary number. Real solutions are tangible and can be represented on a number line.
  • In systems involving quadratics, the nature of the solutions is determined by the discriminant \( b^2 - 4ac \) in the quadratic formula. A positive discriminant indicates two different real solutions, zero indicates exactly one real solution (a repeated root), and a negative discriminant suggests no real solutions, meaning the roots are complex.
  • For the quadratic equation \( 4x^2 + 4x + 1 = 0 \), the discriminant \( 4^2 - 4 \times 4 \times 1 = 0 \) is zero, hence there is one real solution: \( x = -0.5 \).
  • Once we find \( x = -0.5 \), we substitute back to find the corresponding \( y \) value, which is also real, \( y = 1 \), confirming the system has one single real solution: \( (x, y) = (-0.5, 1) \).
Understanding the nature of real solutions helps in comprehending the possibilities and limitations of the solutions of any given system of equations.

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Most popular questions from this chapter

Consider the following system of equations.$$\left\\{\begin{array}{r}x+y=3 \\\\-x+y=1 \\\2 x+y=4\end{array}\right.$$ Use Gauss-Jordan elimination to find the solution, if it exists. Interpret your answer in terms of the graphs of the given equations.

Joi and Cheyenne are planning a party for at least 50 people. They are going to serve hot dogs and hamburgers. Each hamburger costs \(\$ 1\) and each hot dog costs \(\$ .50 .\) Joi thinks that each person will eat only one item, either a hot dog or a hamburger. She also estimates that they will need at least 15 hot dogs and at least 20 hamburgers. How many hamburgers and how many hot dogs should Joi and Cheyenne buy if they want to minimize their cost?

In this set of exercises, you will use the method of solving linear systems using matrices to study real-world problems. Electrical Engineering An electrical circuit consists of three resistors connected in series. The formula for the total resistance \(R\) is given by \(R=R_{1}+R_{2}+R_{3},\) where \(R_{1}, R_{2},\) and \(R_{3}\) are the resistances of the individual resistors. In a circuit with two resistors \(A\) and \(B\) connected in series, the total resistance is 60 ohms. The total resistance when \(B\) and \(C\) are connected in series is 100 ohms. The sum of the resistances of \(B\) and \(C\) is 2.5 times the resistance of \(A\). Find the resistances of \(A, B\), and \(C\).

Involve positive-integer powers of a square matrix \(A . A^{2}\) is defined as the product \(A A ;\) for \(n \geq 3, A^{n}\) is defined as the product \(\left(A^{n-1}\right) A\) For \(n \geq 3\) and a square matrix \(A\), express the inverse of \(A^{n}\) in terms of \(A^{-1} .\)

For the given matrices \(A, B,\) and \(C,\) evaluate the indicated expression. $$\begin{aligned}&A=\left[\begin{array}{rr}3 & 1 \\\2 & 5 \\\\-2 & 1\end{array}\right] ; \quad B=\left[\begin{array}{rr}-5 & -3 \\\1 & 6 \\\8 & 3\end{array}\right]\\\&C=\left[\begin{array}{rrr}2 & 1 & 1 \\\0 & -1 & 7 \\\3 & 0 & -3\end{array}\right] ; \quad C B+2 A\end{aligned}$$
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