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Chapter 11: Problem 14

Solve using the zero-factor property. $$ x^{2}=144 $$

Short Answer

Expert verified
The solutions are \[ x = 12 \] and \[ x = -12 \].

Step by step solution

01

Setting the Equation to Zero

First, we need to ensure the equation is set to zero on one side. The given equation is already in the correct form: \[ x^2 - 144 = 0 \]
02

Factoring the Quadratic Equation

Next, factor the quadratic equation. The equation \[ x^2 - 144 \] can be factored as a difference of squares: \[ (x - 12)(x + 12) = 0 \]
03

Applying the Zero-Factor Property

According to the zero-factor property, if \[ (x - 12)(x + 12) = 0 \], then either \[ x - 12 = 0 \] or \[ x + 12 = 0 \].
04

Solving Each Factor

Solve each equation separately: \[ x - 12 = 0 \] gives \[ x = 12 \]. \[ x + 12 = 0 \] gives \[ x = -12 \].
05

Writing the Solution

The solutions to the equation \[ x^2 = 144 \] are \[ x = 12 \] and \[ x = -12 \].

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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 polynomial equations of degree two. They are in the form \[ ax^2 + bx + c = 0 \], where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. These equations can have two solutions, one solution, or no real solutions depending on the values of \( a \), \( b \), and \( c \). In this exercise, we have a simple quadratic equation \( x^2 = 144 \), which can be solved using the zero-factor property. Understanding quadratic equations is fundamental in algebra as they appear frequently in various mathematical problems. To solve them, we often rely on methods like factoring, completing the square, and the quadratic formula.
Factoring
Factoring is the process of breaking down an equation into simpler terms (factors) that, when multiplied, give the original equation. In the context of quadratic equations, factoring involves finding two binomials whose product equals the quadratic expression. For example, in this exercise, \[ x^2 - 144 \] can be factored into \( (x - 12)(x + 12) \). This method is particularly useful when dealing with quadratic equations because it transforms a complicated equation into a simpler form. Remember, factoring can only be applied when the equation is set to zero on one side as in \[ x^2 - 144 = 0 \].
Difference of Squares
The difference of squares is a specific type of factoring. It applies to expressions in the form \( a^2 - b^2 \). The difference of squares can always be factored into \( (a - b)(a + b) \). This property is what we used in the example problem where \[ x^2 - 144 \] was factored into \( (x - 12)(x + 12) \). It is a powerful technique for simplifying and solving equations because it relies on recognizing patterns in the expressions. By converting a difference of squares into a product of binomials, we make solving for variable values more straightforward.

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Most popular questions from this chapter

Solve for \(x .\) Assume that a and \(b\) represent positive real numbers. \(x^{2}-a^{2}-36=0\)

Use the quadratic formula to solve each equation. (All solutions for these equations are non real complex numbers.) $$ (2 x-1)(8 x-4)=-1 $$

Solve each equation by completing the square. Give (a) exact solutions and (b) solutions rounded to the nearest thousandth. \((x+1)(x+3)=2\)

Find the value of a, b, or c so that each equation will have exactly one rational solution. (Hint: The discriminant must equal 0 for an equation to have one rational solution.) $$ a m^{2}+8 m+1=0 $$

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