Simplify -7 square root of 128(1)x^11(1)y^14
The problem you've presented is a mathematical expression simplification question.
You are required to simplify the expression $-7 square root of 128(1)x^11(1)y^14$, following the mathematical rules for handling square roots, multiplication, and exponents (powers). This typically involves breaking down the square root into its prime factors, reducing it to its simplest radical form, and combining like terms, which, in this case, would mean incorporating the variables $x$and $y$with their respective exponent notations. The final simplified form should no longer have the square root of a non-prime number if possible and should be expressed as a product of a numerical coefficient and variables with simplified exponents.
$- 7 \sqrt{128 \left(\right. 1 \left.\right) x^{11} \left(\right. 1 \left.\right) y^{14}}$
Answer
Solution:
Step 1:
Multiply $128$ by $1$ to maintain the expression. $-7\sqrt{128 \cdot 1 \cdot x^{11} \cdot 1 \cdot y^{14}}$
Step 2:
Simplify the multiplication of $128$ by $1$. $-7\sqrt{128 \cdot x^{11} \cdot y^{14}}$
Step 3:
Express $128 \cdot x^{11} \cdot y^{14}$ as $(8 \cdot x^5 \cdot y^7)^2 \cdot 2 \cdot x$.
Step 3.1:
Extract $64$ from $128$. $-7\sqrt{64 \cdot 2 \cdot x^{11} \cdot y^{14}}$
Step 3.2:
Represent $64$ as $8^2$. $-7\sqrt{8^2 \cdot 2 \cdot x^{11} \cdot y^{14}}$
Step 3.3:
Take out $x^{10}$ from $x^{11}$. $-7\sqrt{8^2 \cdot 2 \cdot (x^{10} \cdot x) \cdot y^{14}}$
Step 3.4:
Rewrite $x^{10}$ as $(x^5)^2$. $-7\sqrt{8^2 \cdot 2 \cdot ((x^5)^2 \cdot x) \cdot y^{14}}$
Step 3.5:
Rewrite $y^{14}$ as $(y^7)^2$. $-7\sqrt{8^2 \cdot 2 \cdot ((x^5)^2 \cdot x) \cdot (y^7)^2}$
Step 3.6:
Rearrange $x$. $-7\sqrt{8^2 \cdot 2 \cdot (x^5)^2 \cdot (y^7)^2 \cdot x}$
Step 3.7:
Rearrange $2$. $-7\sqrt{8^2 \cdot ((x^5)^2) \cdot (y^7)^2 \cdot 2 \cdot x}$
Step 3.8:
Combine terms under the radical. $-7\sqrt{(8 \cdot x^5 \cdot y^7)^2 \cdot 2 \cdot x}$
Step 3.9:
Add parentheses to clarify the expression. $-7\sqrt{((8 \cdot x^5 \cdot y^7)^2) \cdot (2 \cdot x)}$
Step 4:
Extract terms from under the radical, recognizing that the square root of a square is the value itself. $-7(8 \cdot x^5 \cdot y^7 \cdot \sqrt{2 \cdot x})$
Step 5:
Multiply $8$ by $-7$. $-56 \cdot x^5 \cdot y^7 \cdot \sqrt{2 \cdot x}$
Knowledge Notes:
To simplify the given expression, we follow a series of algebraic steps:
Multiplication of Real Numbers: Multiplying real numbers is straightforward; for example, $128 \cdot 1 = 128$.
Square Roots and Exponents: The square root of a number $a^2$ is $a$. Similarly, $\sqrt{x^{2n}} = x^n$ for any integer $n$.
Factoring Perfect Squares: Numbers like $64$ are perfect squares because $64 = 8^2$. This property is used to simplify square roots.
Combining Like Terms: Algebraic expressions with the same variables and exponents can be combined, such as $x^{10} \cdot x = x^{11}$.
Simplifying Radical Expressions: When a term inside a radical is a perfect square, it can be taken out of the radical. For example, $\sqrt{a^2 \cdot b} = a \cdot \sqrt{b}$.
Distributive Property: This property is used when multiplying numbers outside the radical with those inside, such as $-7 \cdot 8 = -56$.
By applying these principles, we can simplify the original expression to its simplest form.
