/*! 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 3 Find all real solutions. $$x^{... [FREE SOLUTION] | 91Ó°ÊÓ

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Find all real solutions. $$x^{4}-x^{2}=2 x^{2}+4$$

Short Answer

Expert verified
The real solutions are \( x = 2 \) and \( x = -2 \).

Step by step solution

01

Write the Equation

First, write down the given equation clearly: \( x^4 - x^2 = 2x^2 + 4 \). This is the equation we're going to solve.
02

Simplify the Equation

Move all terms to one side of the equation to set it equal to zero: \( x^4 - x^2 - 2x^2 - 4 = 0 \). Combine like terms to simplify this to \( x^4 - 3x^2 - 4 = 0 \).
03

Substitute to Form a Quadratic

To make the equation easier to solve, substitute \( y = x^2 \), which transforms the equation into \( y^2 - 3y - 4 = 0 \). This is a quadratic equation in terms of \( y \).
04

Solve the Quadratic Equation

Solve the equation \( y^2 - 3y - 4 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a = 1 \), \( b = -3 \), and \( c = -4 \). Calculate the discriminant: \( b^2 - 4ac = (-3)^2 - 4 \times 1 \times (-4) = 9 + 16 = 25 \). Thus, we have \( y = \frac{3 \pm 5}{2} \).
05

Determine Values for y

Calculate the values for \( y \): \( y = \frac{3 + 5}{2} = 4 \) and \( y = \frac{3 - 5}{2} = -1 \). These are the possible values for \( x^2 \).
06

Find Real Solutions for x

Convert back from \( y \) to \( x \). If \( y = x^2 \), then \( x^2 = 4 \) gives \( x = \pm 2 \). Since \( x^2 = -1 \) does not give real solutions (as \( x^2 \) cannot be negative for real numbers), the only real solutions to the equation are \( x = 2 \) and \( x = -2 \).

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Key Concepts

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

Quadratic Equation
A quadratic equation is a fundamental concept in algebra that deals with polynomial equations of degree two. Such equations have the general form given as: \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Understanding these equations is crucial as they appear frequently in various mathematical and real-world applications.
To solve a quadratic equation, the solutions are typically obtained using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). This formula is derived from completing the square method, a key technique in algebra.
In practice, one often encounters quadratic equations while dealing with physics problems involving projectile motion, in business for calculating break-even points, or in geometry when analyzing figures like circles and parabolas. Recognizing when a problem is a quadratic equation allows you to apply standard methods to find solutions efficiently, which are usually two, considering they are real.
Solving Algebraic Equations
Solving algebraic equations involves finding the values of the variable that make the equation true. Algebra is all about patterns, relationships, and the rules for them. When it comes to solving equations, it often requires isolating the variable on one side.
In our case, to solve the original polynomial equation \( x^4 - x^2 = 2x^2 + 4 \), the essential method was transforming it into a quadratic form by setting \( y = x^2 \). This transformation simplifies the given equation to \( y^2 - 3y - 4 = 0 \), thereby reducing a complex polynomial problem to a manageable quadratic equation.
The process illustrated how strategic substitutions enable solving higher-degree equations by converting them into well-known types, such as quadratics. This makes it easier to apply familiar solutions, especially using the quadratic formula or factoring methods where applicable.
Discriminant Calculations
The discriminant in a quadratic equation provides essential information about the nature and number of solutions for the equation. For the quadratic equation of the form \( ax^2 + bx + c = 0 \), the discriminant \( \Delta \) is calculated as \( b^2 - 4ac \).
This specific value tells us:
  • If \( \Delta > 0 \): There are two distinct real solutions for the equation.
  • If \( \Delta = 0 \): There is exactly one real solution, known as a repeated or double root.
  • If \( \Delta < 0 \): There are no real solutions; instead, there are two complex conjugate solutions.
The discriminant lets us ascertain the solution type even before calculating it. In solving the example quadratic, \( y^2 - 3y - 4 = 0 \), the discriminant \( \Delta = 25 \) implied two distinct real solutions. Consequently, the transformation \( y = x^2 \) led us to the real solutions \( x = 2 \) and \( x = -2 \) for the original equation.

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Most popular questions from this chapter

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