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Chapter 1: Problem 45

Solve by completing the square and applying the square root property. \(t^{2}-8 t=-24\)

Short Answer

Expert verified
The solutions are \( t = 4 + 2\sqrt{10} \) and \( t = 4 - 2\sqrt{10} \).

Step by step solution

01

Move the constant term to the right side

First, add 24 to both sides of the equation to isolate the quadratic and linear terms on the left side. \[ t^2 - 8t + 24 = 0 \] becomes \[ t^2 - 8t = 24 \]
02

Complete the square

To complete the square, add and subtract the square of half the coefficient of the linear term (which is -8) inside the equation. Half of -8 is -4, and its square is 16.Add and subtract 16 on the left-hand side: \[ t^2 - 8t + 16 = 24 + 16 \] This simplifies to \[ (t - 4)^2 = 40 \]
03

Apply the square root property

Take the square root of both sides to solve for t: \[ \sqrt{(t - 4)^2} = \pm \sqrt{40} \] This results in \[ t - 4 = \pm \sqrt{40} \]
04

Simplify the square root

Simplify \(\sqrt{40}\) to \(2\sqrt{10}\): \[ t - 4 = \pm 2\sqrt{10} \]
05

Solve for t

Add 4 to both sides to find the values of t: \[ t = 4 \pm 2\sqrt{10} \]

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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 a type of polynomial equation of degree 2. They have the general form: \[ ax^2 + bx + c = 0 \]where
  • a is the coefficient of the quadratic term
  • b is the coefficient of the linear term
  • c is the constant term
In the given exercise, the quadratic equation is \[ t^2 - 8t = -24 \]Notice that the equation contains a quadratic term \( t^2 \) and a linear term \( -8t \). To solve this quadratic equation using the method of completing the square, it's essential to understand that we need to rearrange it into the form \[ (t - h)^2 = k \]. This transformation allows us to use the square root property to solve for the variable t.
square root property
The square root property is a useful tool for solving equations that have been transformed into a perfect square. It states that if \[ x^2 = k \]then \[ x = \pm \sqrt{k} \]This can also apply to an expression inside the square. For instance, if \[ (t - 4)^2 = 40 \]we can apply the square root property to get: \[ t - 4 = \pm \sqrt{40} \] To solve for t, we'll isolate t by adding 4 to both sides, resulting in:\[ t = 4 \pm \sqrt{40} \]In our given problem, simplifying \( \sqrt{40} \) as \( 2\sqrt{10} \)provides a more refined answer: \[ t = 4 \pm 2\sqrt{10} \]
solving equations
Solving equations often requires moving terms from one side to the other, engaging in operations like addition, subtraction, or applying properties. In completing the square:
  1. We began by isolating the quadratic and linear terms: \[ t^2 - 8t = 24 \]
  2. Next, we needed to create a perfect square trinomial on the left side by adding and subtracting the square of half the coefficient of the linear term (i.e., \( (-8/2)^2 = 16 \) resulting in \[ t^2 - 8t + 16 = 40 \]or \[ (t - 4)^2 = 40 \])This step helps us set up the equation to apply the square root property.
  3. By applying the square root property, we simplified the problem into \[ t - 4 = \pm \sqrt{40} \]Finally, we solved for t by isolating it:\[ t = 4 \pm 2\sqrt{10} \]
Each step ensured that the problem became progressively simpler until reaching the solution. This process is fundamental for solving many types of equations.

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Most popular questions from this chapter

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