Problem 15 - College Algebra

Question

Solve each radical equation in Check all proposed solutions. $$ \sqrt{2 x+13}=x+7 $$

Step-by-Step Solution

Answer

The solution to the radical equation \(\sqrt{2x+13} = x+7\) is \(x = -6\).

Step 1: Isolate the radical

The first step in solving a radical equation is to get the radical on one side of the equation by itself. Here the square root is already isolated, so there is no change in the equation: \[\sqrt{2x+13} = x+7\]

Step 2: Square both sides

Now that the radical term is isolated, remove the square root by squaring both sides of the equation:\[ (\sqrt{2x+13})^2 = (x+7)^2\]This simplifies to:\[2x + 13 = x^2 + 14x + 49\]

Step 3: Rearrange into standard quadratic form

To solve the equation, express it as a quadratic equation in the standard form \(ax^2+bx+c=0\). This is achieved by subtracting \(2x+13\) from both sides:\[0 = x^2 + 12x + 36\]

Step 4: Solve the quadratic equation

This quadratic equation has the form of a perfect square trinomial \(x^2+2bx+b^2\), where \(b=6\). Thus,\[0 = (x+6)^2\]Solving the quadratic equation for \(x\), we get a single solution:\[x = -6\]

Step 5: Checking potential solutions

Since the process of squaring both sides of the equation may introduce extraneous solutions (solutions that mathematically solve the transformed equation but don't actually solve the original problem), we need to plug \(x = -6\) back into the original equation to make sure it truly fits:\[\sqrt{2(-6)+13} = -6 + 7\]which simplifies to:\[1 = 1\]Since the left side equals the right side, \(x = -6\) is a valid solution for the equation.

Key Concepts

Quadratic Equations

When we talk about quadratic equations, we're discussing a fundamental concept in algebra represented generally as \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are constants and \(a\) is not zero. Quadratic equations are known for their U-shaped graph, called a parabola, and they can have zero, one, or two real solutions.

Solving a quadratic means finding the values of \(x\) that make the equation true. There are several methods to solve quadratics: factoring, using the quadratic formula, completing the square, or graphing. In the context of radical equations, quadratics might emerge after squaring both sides to eliminate the radical. A critical skill in solving these is rearranging the equation into the standard quadratic form to identify the most appropriate solving method.

Extraneous Solutions

Extraneous solutions are false or 'extra' solutions that arise when you manipulate an equation—especially when you square both sides of an equation while solving for radicals. This is because squaring both sides does not preserve the 'equivalence' of the original equation, and so while a solution might satisfy the squared equation, it might not satisfy the original one. This is why checking solutions by substituting them back into the original equation is a critical step. If the solution doesn't work in the original equation, it is considered extraneous and should be discarded, leaving only the valid solutions.

Isolate the Radical

Isolating the radical is the primary step in solving radical equations. This involves moving every term except the radical expression to the other side of the equation. Doing this simplifies the equation and prepares it for being squared. It's akin to clearing the stage before a performance; you want the radical expression to be the only actor left, so you can deal with it directly. When the radical is isolated, the next move is typically to apply a power that corresponds to the radical's index to both sides of the equation, which allows for the radical to be eliminated and the equation to be solved by standard algebraic methods.

Perfect Square Trinomial

A perfect square trinomial takes the form \(a^2+2ab+b^2=(a+b)^2\), one of the most delightful patterns to work with in algebra. It's the result of squaring a binomial. In our exercise, after squaring both sides and reorganizing the equation, we identify the quadratic equation \(x^2 + 12x + 36\) as a perfect square trinomial, because it factors neatly into \((x+6)^2\).

The beauty of this format is that solving becomes a much simpler task—you take the square root of both sides, being mindful of the plus and minus possibilities that arise from taking square roots, then solve the subsequent linear equation. It's crucial, however, to check for extraneous solutions since the act of squaring the equation could have introduced them.