Problem 39 - Precalculus : Building Concepts and Connections

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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: \( x^{2}+6x = 16 \).

Step 2: 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 \( x = -8 \).

Step 3: 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.

Key Concepts

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.