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Chapter 1: Problem 51

Solve the equation by factoring. $$2 x^{2}=8$$

Short Answer

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

Step by step solution

01

Rewrite the Equation

We start with the given equation \(2x^2 = 8\). We want all terms on one side of the equation. So, subtract 8 from both sides to get \(2x^2 - 8 = 0\).
02

Factor Out the Greatest Common Factor

Look at every term in the equation \(2x^2 - 8 = 0\). Notice that 2 is a common factor. Factor out 2 from each term, leaving \(2(x^2 - 4) = 0\).
03

Factor the Quadratic Expression

Now, focus on the expression inside the parentheses \(x^2 - 4\). This is a difference of squares, which can be factored as \((x - 2)(x + 2)\). So, we have \(2(x - 2)(x + 2) = 0\).
04

Set Factors to Zero

Apply the zero product property: set each factor equal to zero. So we have two equations: \(x - 2 = 0\) and \(x + 2 = 0\).
05

Solve Each Equation

Solve these simple equations. For \(x - 2 = 0\), add 2 to both sides to get \(x = 2\). For \(x + 2 = 0\), subtract 2 from both sides to get \(x = -2\).

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Key Concepts

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

Quadratic Equations
Understanding quadratic equations is fundamental in algebra. A quadratic equation is typically written in the standard form: \[ ax^2 + bx + c = 0 \]Here the coefficients are
  • \(a\): the coefficient of the term \(x^2\) (not zero),
  • \(b\): the coefficient of the term \(x\),
  • \(c\): the constant term.
Quadratic equations can have:
  • two real solutions,
  • one real solution, or
  • no real solution.
These equations form a parabola when graphed. Solving them often involves methods such as factoring, completing the square, or using the quadratic formula. In the exercise, we used factoring to find the solutions of the equation. By rewriting terms, we utilized factoring to simplify and solve the equation.
Difference of Squares
The concept of difference of squares helps in factoring specific types of quadratic expressions. A difference of squares occurs in expressions like:\[ a^2 - b^2 \]This can be factored into:\[ (a - b)(a + b) \]This pattern arises because multiplying these two binomials,
  • \((a - b)(a + b) = a^2 + ab - ab - b^2\)
yields the original difference of squares, \( a^2 - b^2 \), where the middle terms cancel out. In our example \((x^2 - 4)\), we identify it as \((x^2 - 2^2)\).Thus, it factors into \((x - 2)(x + 2)\).Recognizing and applying the difference of squares can quickly simplify and solve polynomial equations, especially those quadratic in nature.
Zero Product Property
The zero product property is a powerful tool for solving equations. It states that:If\[ ab = 0 \]then either
  • \(a = 0\),
  • \(b = 0\), or
  • both are zero.
This property is essential when working with factored forms of equations. In the solution, after factoring the expression to \(2(x - 2)(x + 2) = 0\),we apply the zero product property by setting each factor equal to zero:
  • \(x - 2 = 0\) gives solution \(x = 2\), and
  • \(x + 2 = 0\) gives solution \(x = -2\).
These roots are found easily, showcasing how this method provides quick and effective ways to find solutions to polynomial equations.
Solving Polynomial Equations
Solving polynomial equations often involves several techniques, but factoring is one of the most straightforward when applicable. The process typically includes:
  • Rewriting the equation so all terms are on one side.
  • Finding a common factor if it exists, and simplifying.
  • Using specific factoring techniques like difference of squares for certain expressions.
  • Applying the zero product property to solve the factored equation.
Each step helps make complex equations more manageable. In our example \(2x^2 = 8\), the steps taken simplify the polynomial into factors that are straightforward to solve. Factoring reduces the equation to the products of simpler expressions, which can then be individually addressed using the zero product property. This method is crucial in algebra, providing clarity and structure in handling complex polynomial equations.

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Most popular questions from this chapter

Factor the expression completely. $$4 x^{2}+4 x y+y^{2}$$

Perform the indicated operations and simplify. $$x^{1 / 4}\left(2 x^{3 / 4}-x^{1 / 4}\right)$$

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Factor the expression completely. Begin by factoring out the lowest power of each common factor. $$x^{5 / 2}-x^{1 / 2}$$

Perform the indicated operations and simplify. $$(x+y+z)(x-y-z)$$
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