/*! 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 39 Solve the radical equation to fi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the radical equation to find all real solutions. Check your solutions. $$\sqrt{x^{2}+6 x}-1=3$$

Short Answer

Expert verified
The solutions to the radical equation are \(x = 2\) and \(x = -8\). Remember to always check your solutions in the original equation.

Step by step solution

01

Eliminate the Square Root

To eliminate the square root, you'll first need to isolate it. Start by adding '1' to both sides of the equation: \( \sqrt{x^{2}+6x}=4 \). Next, square both sides to get rid of the square root: \( x^{2}+6x = 16 \).
02

Solve the Quadratic Equation

Now you have a quadratic equation, which you can solve by first subtracting '16' from both sides to set the equation to zero: \( x^{2}+6x-16=0 \). You can solve this equation through factoring, completing the square or using the quadratic formula. The quadratic formula is \( x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \). In this case, a = 1, b = 6, and c = -16. Using the quadratic formula, you find that \( x = \frac{-6 \pm \sqrt{(6^{2})-4*1*(-16)}}{2*1} = \frac{-6 \pm \sqrt{100}}{2} = \frac{-6 \pm 10}{2} \), so the two solutions are \( x = 2 \) or \( x = -8 \).
03

Check the Solutions

Substitute the found values into the original equation and see if they hold true. For \( x = 2 \), the original equation is \( \sqrt{(2)^{2}+6*2}-1=3 \), which simplifies to \( \sqrt{16}-1=3 \) and finally to \( 4-1=3 \) , which is true. For \( x = -8 \), the original equation is \( \sqrt{(-8)^{2}+6*(-8)}-1=3 \), which simplifies to \( \sqrt{64-48}-1=3 \) and finally to \( 4-1=3 \) , which is also true. So, both \( x = 2 \) and \( x = -8 \) are solutions.

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 91Ó°ÊÓ!

Key Concepts

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

Quadratic Equation
A quadratic equation is a second-order polynomial equation in a single variable x, with a nonzero coefficient for the squared term. It has the general form: \[ax^{2} + bx + c = 0,\] where 'a' is not equal to 0. The solutions to a quadratic equation are called 'roots', and they can be real or complex numbers.

In solving a quadratic equation like \(x^{2}+6x-16=0\), identifying 'a', 'b', and 'c' is the first step. In this case, a = 1, b = 6, and c = -16. This equation is set to equal zero because that allows us to use several methods to find x, including factoring, completing the square, and applying the quadratic formula. Making use of our mathematical toolkit, we aim to uncover the values of x that satisfy the given quadratic relationship.
Square Root Isolation
Isolating the square root in an equation is a strategic first step in solving radical equations.

To isolate the square root, you want to move all other terms to the opposite side of the equation. This serves as a preparation for removing the radical. In the given problem, adding 1 to both sides leads to \(\sqrt{x^{2} + 6x} = 4\).

After isolation, the square root can be eliminated by squaring both sides of the equation. This process, called 'squaring', helps in converting a radical equation into a polynomial one, usually into a quadratic equation as seen in the example. However, it's critical to note that when we square a solution, extra solutions called 'extraneous solutions' may emerge, so verifying each solution in the original equation is an essential step to confirm their validity.
Quadratic Formula
The quadratic formula provides a straightforward way to find the roots of any quadratic equation. It's given by: \[x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\]

This powerful formula comes in handy especially when the quadratic equation does not factor neatly or when completing the square is too cumbersome. Using the given coefficients from the quadratic equation \(x^{2}+6x-16=0\), we plug a=1, b=6, and c=-16 into the formula. This yields: \[x = \frac{-6 \pm \sqrt{6^{2}-4 \cdot 1 \cdot (-16)}}{2 \cdot 1}\] After simplifying under the square root, we arrive at two potential solutions for x.

It's critical to mention that while the quadratic formula always provides the solutions, they must be checked in the original equation to ensure they do not produce negative numbers under the square root or any other contradiction—highlighting the importance of verifying solutions to avoid extraneous results.

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

In Exercises \(101-104,\) let \(f(t)=3 t+1\) and \(g(x)=x^{2}+4\). Evaluate \((g \circ g)\left(\frac{1}{2}\right)\)

In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=|x| ; g(x)=\frac{x^{2}+3}{x^{2}-4}$$

The surface area of a sphere is given by \(A(r)=4 \pi r^{2},\) where \(r\) is in inches and \(A(r)\) is in square inches. The function \(C(x)=6.4516 x\) takes \(x\) square inches as input and outputs the equivalent result in square centimeters. Find \((C \circ A)(r)\) and explain what it represents.

Graph each quadratic function by finding a suitable viewing window with the help of the TABLE feature of a graphing utility. Also find the vertex of the associated parabola using the graphing utility. $$h(x)=(\sqrt{2}) x^{2}+x+1$$

A rectangular garden plot is to be enclosed with a fence on three of its sides and an existing wall on the fourth side. There is 45 feet of fencing material available. (a) Write an equation relating the amount of available fencing material to the lengths of the three sides that are to be fenced. (b) Use the equation in part (a) to write an expression for the width of the enclosed region in terms of its length. (c) For each value of the length given in the following table of possible dimensions for the garden plot, fill in the value of the corresponding width. Use your expression from part (b) and compute the resulting area. What do you observe about the area of the enclosed region as the dimensions of the garden plot are varied? $$\begin{array}{ccc}\hline \begin{array}{c}\text { Length } \\\\\text { (feet) }\end{array} & \begin{array}{c}\text { Width } \\\\\text { (feet) }\end{array} & \begin{array}{c}\text { Total Amount of } \\\\\text { Fencing Material (feet) }\end{array} & \begin{array}{c}\text { Area } \\\\\text { (square feet) }\end{array} \\\\\hline 5 & & 45 \\\10 & & 45 \\\15 & & 45 \\\20 & & 45 \\\30 & & 45 \\\k & & 45 \\\& &\end{array}$$ (d) Write an expression for the area of the garden plot in terms of its length. (e) Find the dimensions that will yield a garden plot with an area of 145 square feet.

See all solutions

Recommended explanations on Math Textbooks

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.