Chapter 11: Problem 37
Solve using the square root property. Simplify all radicals. $$ x^{2}-12=0 $$
Short Answer
Expert verified
x = \pm 2\sqrt{3}
Step by step solution
01
Isolate the variable term
Add 12 to both sides of the equation to isolate the quadratic term: \[ x^2 - 12 + 12 = 12 \]This simplifies to: \[ x^2 = 12 \]
02
Apply the square root property
Take the square root of both sides to solve for \(x\). Remember to consider both the positive and negative roots:\[ \sqrt{x^2} = \pm \sqrt{12} \]This simplifies to:\[ x = \pm \sqrt{12} \]
03
Simplify the radical
Express \( \sqrt{12} \) in its simplest form by factoring out the perfect square:\[ \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3} \]Thus, the solutions are: \[ x = \pm 2\sqrt{3} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
square root property
The square root property is a useful tool in solving quadratic equations. Essentially, it states that if you have an equation of the form \( x^2 = k \), you can solve for \( x \) by taking the square root of both sides. This results in: \( x = \pm \sqrt{k} \). In mathematical notation, \( \pm \) means 'plus or minus'. This signifies that there are always two solutions: a positive root and a negative root.
For example, in the equation \( x^2 = 12 \), applying the square root property gives: \( x = \pm \sqrt{12} \). Remember, you must always consider both positive and negative possibilities. This ensures you capture all potential solutions to the equation.
Applying the square root property is straightforward:
For example, in the equation \( x^2 = 12 \), applying the square root property gives: \( x = \pm \sqrt{12} \). Remember, you must always consider both positive and negative possibilities. This ensures you capture all potential solutions to the equation.
Applying the square root property is straightforward:
- Isolate the squared term (move other terms to the opposite side).
- Take the square root of both sides of the equation.
- Include both positive and negative roots in your solutions.
simplifying radicals
Simplifying radicals involves breaking down a radical expression into its simplest form. This makes the expression easier to understand and use in further calculations. To simplify a radical like \( \sqrt{12} \):
This method ensures your solutions are in their simplest and most understandable form.
- First, factor the number under the radical into its prime factors. For 12, you get: \(12 = 4 \times 3 \).
- Next, find the perfect squares within those factors. Here, \( 4 \) is a perfect square since \( \sqrt{4} = 2 \).
- Separate the perfect square from the other factors. So, \( \sqrt{12} = \sqrt{4 \times 3} \).
- Simplify by taking the square root of the perfect square part. This gives: \( \sqrt{4} \times \sqrt{3} = 2 \sqrt{3} \).
This method ensures your solutions are in their simplest and most understandable form.
factoring perfect squares
Factoring perfect squares means rewriting a squared term as a product of its square root. This technique is often used to simplify equations and expressions.
Consider a quadratic like \( x^2 - 12 \). Although 12 is not a perfect square, let's say you had \( x^2 - 16 \). You could rewrite 16 as \( 4^2 \), a perfect square. This simplifies to: \( x^2 - 4^2 \). Now, apply the difference of squares formula: \( x^2 - a^2 = (x - a)(x + a) \).
In this case, \( x^2 - 4^2 = (x - 4)(x + 4) \).
Perfect square trinomials also use this concept. For example, \( x^2 + 6x + 9 \) can be factored as: \( (x + 3)^2 \), because: - The first term is a square: \( x^2 \). - The last term is a square: \( 9 = 3^2 \). - The middle term is twice the product of the first and last terms’ square roots: \( 2 \times x \times 3 = 6x \). The better you get at recognizing perfect squares, the quicker you can factor and solve quadratic problems.
Consider a quadratic like \( x^2 - 12 \). Although 12 is not a perfect square, let's say you had \( x^2 - 16 \). You could rewrite 16 as \( 4^2 \), a perfect square. This simplifies to: \( x^2 - 4^2 \). Now, apply the difference of squares formula: \( x^2 - a^2 = (x - a)(x + a) \).
In this case, \( x^2 - 4^2 = (x - 4)(x + 4) \).
Perfect square trinomials also use this concept. For example, \( x^2 + 6x + 9 \) can be factored as: \( (x + 3)^2 \), because: - The first term is a square: \( x^2 \). - The last term is a square: \( 9 = 3^2 \). - The middle term is twice the product of the first and last terms’ square roots: \( 2 \times x \times 3 = 6x \). The better you get at recognizing perfect squares, the quicker you can factor and solve quadratic problems.