/*! 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 36 \(x^{2}-x-12=0\)... [FREE SOLUTION] | 91Ó°ÊÓ

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\(x^{2}-x-12=0\)

Short Answer

Expert verified
The solutions are \(x = 4\) and \(x = -3\).

Step by step solution

01

Factoring the Quadratic Equation

The equation we have is a quadratic equation: \[ x^2 - x - 12 = 0 \]We need to factor the quadratic expression. We will look for two numbers that multiply to \(-12\) (the constant term) and add up to \(-1\) (the coefficient of the \(x\) term). These numbers are \(-4\) and \(+3\).
02

Rewriting the Quadratic as a Factored Expression

Using the numbers found in Step 1, rewrite the quadratic expression:\[ (x - 4)(x + 3) = 0 \]This step transforms the quadratic equation into a product of two binomials.
03

Solving Each Factor for Zero

To find the values of \(x\) that make the equation true, set each factor equal to zero and solve:1. \(x - 4 = 0\): \[ x = 4 \]2. \(x + 3 = 0\): \[ x = -3 \]
04

Verifying the Solutions

Ensure that both solutions solve the original equation. Substitute \(x = 4\) and \(x = -3\) into \(x^2 - x - 12 = 0\).1. For \(x = 4\): \[ 4^2 - 4 - 12 = 16 - 4 - 12 = 0 \]2. For \(x = -3\): \[ (-3)^2 - (-3) - 12 = 9 + 3 - 12 = 0 \]Both solutions satisfy the equation.

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Key Concepts

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

Factoring a Quadratic Equation
Factoring is a method used to solve quadratic equations by expressing them as a product of simpler expressions. For the quadratic equation \( x^2 - x - 12 = 0 \), we need to factor the expression \( x^2 - x - 12 \). Our goal is to identify two numbers that multiply to \(-12\) (the constant term) and add up to \(-1\) (the coefficient of the \(x\) term). These magic numbers are \(-4\) and \(+3\). When we use these numbers, the quadratic can be rewritten in a factored form as:
  • \((x - 4)(x + 3) = 0\)
This new expression means the equation is now the product of two binomials. Factoring helps simplify the equation, making it easier to solve for the variable \(x\). Always check your factors to ensure they indeed satisfy the original terms of multiplication and addition.
Finding the Roots of the Quadratic Equation
Once the quadratic equation is factored, the next step is finding the roots. Roots are the values of \(x\) that satisfy the quadratic equation, meaning they make the equation true (equal to zero). For the factored equation \((x - 4)(x + 3) = 0\), each binomial expression must be set to zero:
  • First, solve \(x - 4 = 0\). This gives the root \(x = 4\).
  • Next, solve \(x + 3 = 0\). This gives the root \(x = -3\).
By solving these simple equations, we identify the roots of the quadratic equation as \(x = 4\) and \(x = -3\). These roots are crucial because they represent points where the parabola described by the quadratic equation cuts the x-axis.
Verification of Solutions
Verification is a key step in problem-solving, ensuring the solution is correct. After finding the roots of the quadratic equation \((x = 4\) and \(x = -3\)), we verify them by substituting back into the original equation \(x^2 - x - 12 = 0\).
  • For \(x = 4\): Substituting gives \(4^2 - 4 - 12 = 16 - 4 - 12 = 0\), satisfying the equation.
  • For \(x = -3\): Substituting gives \((-3)^2 - (-3) - 12 = 9 + 3 - 12 = 0\), also satisfying the equation.
These calculations confirm that both proposed solutions \(x = 4\) and \(x = -3\) are indeed correct. This step is crucial in mathematics to ensure no errors have been made during the factoring and root-finding processes.

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Most popular questions from this chapter

$$ \text { For Problems } 41-70 \text {, solve each equation. (Objective } 3 \text { ) } $$ $$ -2 x^{3}+8 x=0 $$

Suppose that the width of a certain rectangle is 3 inches less than its length. The area is numerically 6 less than twice the perimeter. Find the length and width of the rectangle.

$$ \text { For Problems } 41-70 \text {, solve each equation. (Objective } 3 \text { ) } $$ $$ -3 x^{2}-24 x=0 $$

For Problems \(71-88\), set up an equation and solve each problem. (Objective 4) Find two numbers whose product is \(-1\). One of the numbers is three more than twice the other number.

Suppose that we want to factor \(n^{2}+26 n+168\) so that we can solve the equation \(n^{2}+26 n+168=0\). We need to find two positive integers whose product is 168 and whose sum is 26 . Since the constant term, 168, is rather large, let's look at it in prime factored form: $$ 168=2 \cdot 2 \cdot 2 \cdot 3 \cdot 7 $$ Now we can mentally form two numbers by using all of these factors in different combinations. Using two 2 s and the 3 in one number and the other 2 and the 7 in another number produces \(2 \cdot 2 \cdot 3=12\) and \(2 \cdot 7=14\). Therefore, we can solve the given equation as follows: $$ \begin{aligned} n^{2}+26 n+168 &=0 & \\ (n+12)(n+14)=0 & & \\ n+12=0 & \text { or } & n+14 &=0 \\ n=-12 & \text { or } & n &=-14 \end{aligned} $$ The solution set is \(\\{-14,-12\\}\). Solve each of the following equations. (a) \(n^{2}+30 n+216=0\) (b) \(n^{2}+35 n+294=0\) (c) \(n^{2}-40 n+384=0\) (d) \(n^{2}-40 n+375=0\) (e) \(n^{2}+6 n-432=0\)

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