Problem 36 - 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. $$x^{2}-10 x+25=3$$

Step-by-Step Solution

Answer

The solutions for the equation \(x^{2} - 10x + 25 = 3\) are \(x = 5 + \sqrt{3}\) and \(x = 5 - \sqrt{3}\).

Step 1: Factor the Perfect Square Trinomial

The quadratic on the left side can be factored into \((x-5)^{2} = 3.\)

Step 2: Apply the Square Root Property

Then we apply the square root property to both sides which gives us \(x - 5 = \pm \sqrt{3}.\)

Step 3: Simplify the Equation

After applying the square root property, the next step is to simplify the equation which gives \(x = 5 \pm \sqrt{3}\).

Key Concepts

Factoring Perfect Square Trinomials

When we come across a quadratic equation that looks like a perfect square trinomial, our goal is to factor it in such a way that it will greatly simplify the problem. A perfect square trinomial takes the form of \(a^2 \pm 2ab + b^2\), where the factored form is \(a \pm b)^2\). It's essentially squaring a binomial. To factor such an equation, identify two numbers that multiply to get the constant term (in our example, 25) and also add up to the middle coefficient (in our example, -10).

Once you've identified these two numbers, which both happen to be 5 in our exercise, the equation can be rewritten to highlight the squared binomial, thus \(x^2 - 10x + 25\) becomes \(x - 5)^2\). Recognizing this pattern can save you a lot of time and energy in solving quadratic equations and is an essential skill for algebra students.

Square Root Property

The square root property is a valuable tool in algebra, especially when solving quadratic equations that have been factored into a perfect square. It states that if \(a^2 = b\), then \(a = \pm\sqrt{b}\). This is because squaring a positive or a negative number both give you a positive result, so you must consider both possibilities when taking the square root.

In our example, after factoring the perfect square trinomial, we obtain \(x - 5)^2 = 3\). Applying the square root property to both sides leads us to \(x - 5 = \pm\sqrt{3}\), effectively breaking down the equation into two simpler linear equations to solve for \(x\). This step is critical in finding the final solutions.

Simplifying Radicals

Radicals may seem daunting at first, but simplifying them is just about recognizing patterns and applying elementary division. The goal when simplifying radicals is to express the square root of a number as simply as possible. For numbers that are perfect squares, like \(4\), \(9\), or \(16\), this is straightforward because \(\sqrt{4} = 2\), \(\sqrt{9} = 3\), and \(\sqrt{16} = 4\).

However, when dealing with numbers that aren't perfect squares, like \(3\) in our exercise, we can't simplify \(\sqrt{3}\) any further since \(3\) is a prime number. Therefore, when we reach the final step of our quadratic equation, \(x = 5 \pm\sqrt{3}\) remains as is. In more complex scenarios, simplifying radicals could involve finding and extracting square factors from within a radical or rationalizing the denominator when dealing with fractions.