/*! 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 35 Solve using the square root prop... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 35

Solve using the square root property. $$(p+3)^{2}+4=2$$

Short Answer

Expert verified
The solution for the given equation \((p+3)^2 + 4 = 2\) is \(p = -3 \pm \sqrt{-2}\).

Step by step solution

01

Rewrite the equation by removing 4 from both sides

Subtract 4 from both sides of the equation to isolate the squared term on one side. \[(p+3)^2 + 4 - 4 = 2 - 4\] This simplifies the equation to: \[(p+3)^2 = -2\]
02

Apply the square root property to solve for p

Use the square root property, which states that if \(x^2 = k\), then \(x = \pm \sqrt{k}\). Apply this property to our equation \((p+3)^2 = -2\): \[p+3 = \pm \sqrt{-2}\]
03

Solve for p

Subtract 3 from both sides of the equation to isolate p: \[p = -3 \pm \sqrt{-2}\] The solution for the given equation is: \[p = -3 \pm \sqrt{-2}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Square Root Property
The square root property is a handy method for solving quadratic equations of the form
  • When you see an equation such as \((x - a)^2 = k\), you can take the square root of both sides to simplify things. By doing so, you end up with two possible solutions: \(x - a = \pm \sqrt{k}\).
  • In the exercise, we used this method on \((p+3)^2 = -2\). To apply the square root property, we consider both the positive and negative roots. Thus, we say: \(p+3 = \pm \sqrt{-2}\).
  • It's crucial to remember that the square root of a negative number brings us to complex numbers, an essential algebra concept. Here, it鈥檚 not just about finding one solution, but understanding there are indeed two possible expressions.
The square root property is not just about numbers but understanding how one transformation opens an avenue for solution.
Complex Numbers in Algebra
Complex numbers often pop up when equations lead you to take the square root of a negative number. This may initially seem strange, but in mathematics, every problem has a logical solution. Complex numbers have two components: real and imaginary.
  • The imaginary unit is \(i\), which by definition, equals \(\sqrt{-1}\). Thus, the square root of any negative number can be expressed in terms of \(i\).
  • In our equation, \(\sqrt{-2}\) translates to \(i\sqrt{2}\). This conversion helps by giving us the equation: \[p = -3 \pm i\sqrt{2}\].
  • This shows the beauty and consistency of algebra: even if a number doesn't appear real initially, by expressing it with complex numbers, it becomes manageable and clear.
These numbers might seem abstract, but they have practical applications in fields like engineering, physics, and more.
Quadratic Equation Solutions
Solving quadratic equations doesn't always lead you to straightforward numbers. Sometimes, like in the exercise, you find yourself dealing with a nice blend of algebraic knowledge.
  • The original problem, which had a quadratic term \((p+3)^2\), got simplified using the square root property.
  • Given that the equation led to a negative square, illustrating the presence of complex numbers was the natural progression.
  • In algebra, solutions can then be written as: \[p = -3 \pm \sqrt{-2}\], which translates to \[p = -3 \pm i\sqrt{2}\]. This indicates there are two distinct roots, albeit both involving complex numbers.
This exercise is a perfect match of various algebraic techniques that remind us of the depth and breadth of solving quadratic equations. You encounter real and imaginary parts that together hold practical significance in the broader spectrum of algebra.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write an equation and solve. Chandra cuts fabric into isosceles triangles for a quilt. The height of each triangle is 1 in. less than the length of the base. The area of each triangle is 15 in \(^{2}\). Find the height and base of each triangle.

Solve. 5(t+9)^{2}+37(t+9)+14=0

SOLVE. \(z^{2}+3=0\)

The illuminance \(E\) (measure of the light emitted, in lux ) of a light source is given by $$E=\frac{I}{d^{2}}$$ where \(I\) is the luminous intensity (measured in candela) and \(d\) is the distance, in meters, from the light source. The luminous intensity of a lamp is 2700 candela at a distance of \(3 \mathrm{m}\) from the lamp. Find the illuminance, \(E,\) in lux.

Write an equation and solve. A rectangular swimming pool is 60 ft wide and \(80 \mathrm{ft}\) long. A nonskid surface of uniform width is to be installed around the pool. If there is \(576 \mathrm{ft}^{2}\) of the nonskid material, how wide can the strip of the nonskid surface be?
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.