Problem 19
Question
For the following exercises, solve the quadratic equation by using the square root property. $$ x^{2}=36 $$
Step-by-Step Solution
Verified Answer
The solutions are \(x = 6\) and \(x = -6\).
1Step 1: Understand the Square Root Property
The square root property states that if you have an equation in the form \(x^2 = k\), then \(x = \pm \sqrt{k}\). This property can be used to solve quadratic equations quickly when they are in this format.
2Step 2: Apply the Square Root Property
Given the equation \(x^2 = 36\), identify \(k\) as 36. Apply the square root property by taking the square root of both sides of the equation. This gives us \(x = \pm \sqrt{36}\).
3Step 3: Simplify the Square Root
Calculate the square root of 36. Since \(\sqrt{36} = 6\), the equation becomes \(x = \pm 6\). This means that the solutions for \(x\) are both positive 6 and negative 6.
Key Concepts
Square Root PropertySolving QuadraticsSimplifying Radicals
Square Root Property
The square root property is a quick and powerful tool for solving quadratic equations of the form \(x^2 = k\). This property states that you can find the variable \(x\) by taking the square root of both sides of the equation.
When you take the square root of \(k\), remember to include both the positive and negative roots. This is because squaring either a positive or a negative number will result in a positive product. Hence, for an equation like \(x^2 = 36\), we apply the square root property to get \(x = \pm \sqrt{36}\). This results in two possible solutions for \(x\).
Using this method, you quickly solve for \(x\) without going through a lengthy quadratic formula process, making it a handy approach in many situations.
When you take the square root of \(k\), remember to include both the positive and negative roots. This is because squaring either a positive or a negative number will result in a positive product. Hence, for an equation like \(x^2 = 36\), we apply the square root property to get \(x = \pm \sqrt{36}\). This results in two possible solutions for \(x\).
Using this method, you quickly solve for \(x\) without going through a lengthy quadratic formula process, making it a handy approach in many situations.
Solving Quadratics
Solving quadratic equations involves finding the values of \(x\) that make the equation true. There are several methods to solve quadratic equations, but the square root property is particularly effective when the equation is in the format \(x^2 = k\). Other methods include factoring, completing the square, and using the quadratic formula.
- Factoring involves rewriting the quadratic in a factorable form, though it doesn't always work if the quadratic is not factorable over the integers.
- Completing the square is a method that involves rearranging the equation into a perfect square trinomial, making it solvable by extracting its square root.
- The quadratic formula is a universal method that can solve any quadratic equation, though it can be more complex than simply using the square root property for equations like \(x^2 = k\).
Simplifying Radicals
Simplifying radicals is a crucial step when using the square root property because it helps to express the solution in its simplest form. A radical is an expression that involves roots, such as square roots or cube roots. In cases where the radicand, which is the number under the radical sign, is a perfect square, you can simplify the radical by taking the square root.
Let's take our given exercise, \(x^2 = 36\). Here, \(\sqrt{36}\) is simplified to 6 because 36 is a perfect square (since \(6 \times 6 = 36\)). Therefore, the solutions \(x = \pm \sqrt{36}\) simplify to \(x = \pm 6\). In more complex cases, where the radicand is not a perfect square, you would express it in terms of nearest perfect squares or simplified radical forms.
Breaking down radicals into their simplest form makes the solutions cleaner and easier to interpret, which is valuable for clear communication in math.
Let's take our given exercise, \(x^2 = 36\). Here, \(\sqrt{36}\) is simplified to 6 because 36 is a perfect square (since \(6 \times 6 = 36\)). Therefore, the solutions \(x = \pm \sqrt{36}\) simplify to \(x = \pm 6\). In more complex cases, where the radicand is not a perfect square, you would express it in terms of nearest perfect squares or simplified radical forms.
Breaking down radicals into their simplest form makes the solutions cleaner and easier to interpret, which is valuable for clear communication in math.
Other exercises in this chapter
Problem 19
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