Problem 37 - Introductory Algebra for College Students

Question

Solve each quadratic equation by first factoring the perfect square trinomial on the left side. Then apply the square root property. Simplify radicals, if possible. $$x^{2}+2 x+1=5$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x=-1+\sqrt{5}\) and \(x=-1-\sqrt{5}\).

Step 1: Factoring the perfect square trinomial

The trinomial \(x^{2}+2 x+1\) is a perfect square trinomial. It factors into \((x+1)^2\).

Step 2: Applying the Square Root Property

Given the factored form from step 1, the equation becomes \((x+1)^2=5\). The square root property states that if \(p^2=q\), then \(p=±\sqrt{q}\). Applying this property, we have \(x+1=±\sqrt{5}\).

Step 3: Solving for x

Subtract 1 from both sides of the equation to isolate \(x\), so \(x=-1±\sqrt{5}\).

Key Concepts

Factoring Perfect Square Trinomials

When we encounter a quadratic equation in the form of a perfect square trinomial, we're dealing with something that can be simplified into the square of a binomial. But what exactly is a perfect square trinomial? It's an expression of the form \(ax^2 + 2abx + b^2\), which can be factored into \( (ax + b)^2\). The essential characteristic here is that the first and last terms are perfect squares, and the middle term is twice the product of the square roots of these perfect squares.

Doing this factoring simplifies the problem significantly because we can then apply the square root property in the following steps. For example, the equation given in the exercise, \(x^2 + 2x + 1\), is a textbook case. Each term is a perfect square, and the middle term is twice the product of 1 (the square root of \(x^2\)) and 1 (the square root of 1). Thus, the factored form is \( (x + 1)^2\), setting us up to further solve the equation using other properties.

Square Root Property

Moving forward to the square root property, it's a useful tool for solving equations where a variable is squared. The square root property tells us that if we have an equation in the form of \( p^2 = q\), then the solution(s) for \( p\) can be expressed as \( p = ±\sqrt{q}\). The ± sign indicates that both the positive and negative square roots are solutions to the equation.

This property is particularly handy after factoring a quadratic equation into a perfect square trinomial, as we did in the first step of our example.

Square Root Application

As we applied the property to the factored equation \( (x+1)^2 = 5\), we derived the solutions \(x+1 = ±\root{q}{5}\). By incorporating this property, we transformed the problem into a much simpler and more manageable form, with the remaining task being to solve for \( x\).

Simplifying Radicals

Now let's tackle the last step: simplifying radicals. After applying the square root property, we may end up with a radical expression. Simplifying radicals means breaking down the radical into its simplest form. Sometimes this involves finding the largest square factor of the number inside the radical and removing it from under the radical sign, if possible.

However, in the case of the equation we're solving, \(±\root{q}{5}\) already is in its simplest radical form because 5 is a prime number and does not have any square factors other than 1. So there's no further simplification necessary, and we can proceed to solve for \(x\) by isolating it: \(x = -1±\root{q}{5}\). It is crucial for students to recognize when a radical is already simplified to avoid unnecessary steps and to accurately find the solution to the problem at hand.