Ferdinand

University teacher - Tutor for 7 years

MARCH 14, 2024

The square root property is a pivotal concept in algebra for solving quadratic equations. The square root property simplifies the process of finding the roots of a quadratic equation. It is particularly useful because it gives a direct way to solve for $ x $ when the equation is already in the form of $ x^2 = a $, without the need for more complex methods.

The properties of square roots include:

**Product Property**: $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$, where $a \geq 0$ and $b \geq 0$.**Quotient Property**: $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$, where $a \geq 0$ and $b > 0$.**Square Root of a Square**: $\sqrt{x^2} = |x|$, reflecting that the square root returns the non-negative root.**Rationalizing the Denominator**: Involves multiplying top and bottom by a term that eliminates the square root in the denominator.**Even Roots of Negative Numbers**: Not defined in the real number system. $√(-a)$ is imaginary if $a > 0$.

**Detailed Steps to Solve Quadratic Equations Using the Square Root Property:**

**Converting the equation:**Convert the equation into the form that conforms to the square root Property.**Identify the Numbers**: Determine the values of a, b, and x.**Apply the Square Root**: Apply the square root to both sides of the equation, keeping in mind to include the positive and negative roots.**Simplify the Equation**: After applying the square root property, ensure the other side of the equation is a simplified constant. If there are other terms, add or subtract them.**Solve for $ x $**: After simplifying the equation, you will get two possible solutions for $ x $: $ x = +\sqrt{a} $ and $ x = -\sqrt{a} $.**Check Your Solutions**: It's good practice to substitute your solutions back into the original equation to ensure they work.

Multiply $\sqrt{12}$ and $\sqrt{3}$.

Step 1: Identify the Numbers

- $a = 12$
- $b = 3$

Step 2: Apply the Product Property

- $\sqrt{12} \cdot \sqrt{3} = \sqrt{12 \cdot 3}$

Step 3: Calculate the Product

- $12 \cdot 3 = 36$

Step 4: Simplify the Square Root

- $\sqrt{36} = 6$

Solution:

- $\sqrt{12} \cdot \sqrt{3} = 6$

Simplify $\sqrt{\frac{50}{2}}$.

Step 1: Identify the Numbers

- $a = 50$
- $b = 2$

Step 2: Apply the Quotient Property

- $\sqrt{\frac{50}{2}} = \frac{\sqrt{50}}{\sqrt{2}}$

Step 3: Simplify Each Square Root (if possible)

- $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$ (since $\sqrt{25} = 5$)
- $\sqrt{2}$ remains as it is because it cannot be simplified further.

Step 4: Form the Simplified Expression

- $\frac{5\sqrt{2}}{\sqrt{2}}$

Step 5: Simplify the Expression

- When you divide $5\sqrt{2}$ by $\sqrt{2}$, the $\sqrt{2}$ in the numerator and denominator cancel out, leaving $5$.

Solution:

- $\sqrt{\frac{50}{2}} = 5$

Evaluate $\sqrt{(-5)^2}$.

Step 1: Square the Number

- $(-5)^2 = 25$.

Step 2: Apply the Square Root

- $\sqrt{25}$.

Step 3: Evaluate the Square Root

- $\sqrt{25} = 5$.

Step 4: Apply the Property

- Since $\sqrt{(-5)^2}$ results in $5$, but knowing $-5$ was the original number, we acknowledge $\sqrt{(-5)^2} = |{-5}|$.

Conclusion:

- $\sqrt{(-5)^2} = |{-5}| = 5$, which matches the result of step 3, confirming the property $\sqrt{x^2} = |x|$.

Let's rationalize the denominator for the expression $\frac{3}{\sqrt{2}}$.

Step 1: Identify the Expression

- Fraction with a square root in the denominator: $\frac{3}{\sqrt{2}}$.

Step 2: Multiply by a Form of 1

- To rationalize, multiply both the numerator and denominator by the square root in the denominator: $\sqrt{2}$.
- Form of 1: $\frac{\sqrt{2}}{\sqrt{2}}$.

Step 3: Apply the Multiplication

- Multiply the numerator: $3 \times \sqrt{2} = 3\sqrt{2}$.
- Multiply the denominator: $\sqrt{2} \times \sqrt{2} = 2$.

Step 4: Write the Simplified Expression

- The resulting expression: $\frac{3\sqrt{2}}{2}$.

Conclusion:

- The expression $\frac{3}{\sqrt{2}}$ is rationalized as $\frac{3\sqrt{2}}{2}$.

Find the square root of $-16$.

Step 1: Express $-16$ as $16 \times -1$

- $-16 = 16 \times -1$.

Step 2: Apply the Square Root

- Since the square root of $-1$ is $i$, and the square root of $16$ is $4$, you take the square root of each factor separately:
- $\sqrt{-16} = \sqrt{16} \times \sqrt{-1}$.

Step 3: Simplify Using the Definition of $i$

- $\sqrt{16} = 4$ and $\sqrt{-1} = i$.

Step 4: Combine the Results

- $4 \times i = 4i$.

Conclusion:

- The square root of $-16$ is $4i$, illustrating how to handle even roots of negative numbers by using imaginary numbers.

A: The difference between square and square root is:

**Square**: Multiplying a number by itself. Expressed as $n^2$. For $n=3$, square is $3^2 = 9$.

**Square Root**: Finding a number that, when multiplied by itself, gives the original number. Expressed as $\sqrt{n}$. For $n=9$, square root is $\sqrt{9} = 3$.

A: The vocabulary of square root includes several key terms:

**Square Root**: The square root of a number $x$ is a value that, when multiplied by itself, equals $x$. For example, the square root of 9 ($\sqrt{9}$) is 3, because $3 \times 3 = 9$.**Radicand**: The number under the square root symbol. In $\sqrt{9}$, 9 is the radicand.**Radical Sign**: The symbol $\sqrt{}$ used to denote the square root. For example, in $\sqrt{16}$, the symbol $\sqrt{}$, is the radical sign.**Principal Square Root**: The non-negative square root of a number. Every positive number has two square roots: a positive and a negative. The principal square root refers to the positive one. For example, the principal square root of 16 is 4, not -4.**Perfect Square**: A number that is the square of an integer. For instance, 16 is a perfect square because it’s $4^2$.**Rationalize**: The process of adjusting an expression to eliminate square roots from the denominator. For example, to rationalize $\frac{1}{\sqrt{2}}$, you multiply both numerator and denominator by $\sqrt{2}$ to get $\frac{\sqrt{2}}{2}$.**Imaginary Unit ($i$)**: Represented by $i$, it is defined as $\sqrt{-1}$. It’s used to express square roots of negative numbers, given that no real number squared equals a negative number. For instance, $\sqrt{-4} = 2i$.

A: A number that is both a perfect square and a perfect cube is called a perfect sixth power. For example, $64$ is both a perfect square ($8^2$) and a perfect cube ($4^3$).

A: A number is irrational if it meets certain criteria, indicating it cannot be exactly written as a ratio of two integers $a/b$, where $a$ and $b$ are integers and $b \neq 0$. Key characteristics of irrational numbers include:

**Non-Terminating, Non-Repeating Decimal Expansion**: Irrational numbers have infinite decimal digits that do not terminate or form a repeating pattern.**Cannot Be Expressed as a Fraction**: Unlike rational numbers, which can be written as fractions, irrational numbers cannot be precisely represented in fractional form.**Roots of Non-Perfect Squares (and other degrees)**: The square roots (and roots of other degrees) of non-perfect squares are often irrational. For example, $\sqrt{2}$ cannot be represented as a finite or repeating decimal, nor as a fraction of two integers, making it irrational.

**Example**: Proving $\sqrt{2}$ is Irrational

To understand why $\sqrt{2}$ is irrational, consider if it were rational, it could be written as $\sqrt{2} = \frac{a}{b}$, where $a$ and $b$ are coprime integers (their greatest common divisor is 1), and $b \neq 0$.

- Squaring both sides yields $2 = \frac{a^2}{b^2}$, so $a^2 = 2b^2$.
- This equation implies $a^2$ is even (since it's 2 times some number).
- If $a^2$ is even, then $a$ itself must be even (since the square of an odd number is odd).
- Let $a = 2k$ (where $k$ is an integer), then substituting back: $4k^2 = 2b^2$ or $2k^2 = b^2$, showing $b^2$, and thus $b$, must also be even.
- However, this contradicts our assumption that $a$ and $b$ are coprime, as both being even means they share at least a factor of $2$.

Therefore, no such integers $a$ and $b$ exist that can satisfy the condition $\sqrt{2} = \frac{a}{b}$, making $\sqrt{2}$ irrational. This method illustrates a fundamental way of identifying an irrational number through contradiction.