Problem 110
Question
Explain why the given statement is true for any numbers \(c\) and \(d .\) I Hint: Look at the properties of absolute value on page 10.1 $$\sqrt{9 c^{2}-18 c d+9 d^{2}}=3|c-d|$$
Step-by-Step Solution
Verified Answer
Question: Prove the following statement for any numbers c and d: \(|3c - 3d| = \sqrt{9c^2 - 18cd + 9d^2}\).
Solution: By following the steps outlined above, we demonstrated that the given statement is true for any numbers c and d. The left side of the equation, \(|3c - 3d|\), simplifies to \(3|c - d|\), which is equal to the right side of the equation, thus proving the statement.
1Step 1: Expand the expression under the square root
Start by expanding the expression in the square root:
$$9c^2 - 18cd + 9d^2 = (3c-3d)^2$$
Notice that we factored this expression as a perfect square: \((3c-3d)^2\).
2Step 2: Take the square root of both sides
Now take the square root of both sides:
$$\sqrt{(3c-3d)^2} = \sqrt{9c^2 - 18cd + 9d^2}$$
Since the square root and the square functions are inverse operations, the left side simplifies as follows:
$$3c - 3d = \sqrt{9c^2 - 18cd + 9d^2}$$
3Step 3: Calculate the absolute value of the difference
We are now interested in the absolute value of the left side. Let's calculate the absolute value of the difference \(3c - 3d\):
$$|3c - 3d| = 3|c - d|$$
Here we have factored out the constant 3 from the absolute value, as it is always positive, and it doesn't affect the absolute value of the expression.
4Step 4: Final comparison
Finally, we'll compare our result to the right side of the given equation:
$$3|c - d| = 3|c - d|$$
The LS and the RS are equal, so we have proven that for any numbers \(c\) and \(d\), the given statement holds true.
Key Concepts
Factoring Perfect SquaresInverse OperationsSquare Root PropertiesExpression Simplification
Factoring Perfect Squares
Factoring perfect squares is a useful technique in algebra that simplifies expressions. It's like reverse engineering an expression back into its multiplied form. In our example, we started with the expression \(9c^2 - 18cd + 9d^2\). By noticing the pattern, we factored it into \((3c - 3d)^2\). To do this:
- Recognize patterns that resemble \((a - b)^2 \equiv a^2 - 2ab + b^2\).
- In this case, \(9c^2 = (3c)^2\), \(9d^2 = (3d)^2\), and \(-18cd = -2 \times 3c \times 3d\).
Inverse Operations
Inverse operations are mathematical operations that reverse the effect of each other. The square and square root functions are perfect examples. When we have a squared expression, the square root operation brings it back to its original base. In the solution:
- We started with \((3c-3d)^2\) and applied the square root to both sides.
- Thus, \(\sqrt{(3c-3d)^2} = 3c-3d\).
Square Root Properties
Square root properties are key to understanding how expressions simplify. They tell us that the square root of a squared number is essentially the absolute value of the original number or expression. This holds because:
- For any expression \(x\), \(\sqrt{x^2} = |x|\).
- In our case, after factoring, \(\sqrt{9c^2 - 18cd + 9d^2} = |3c - 3d|\).
Expression Simplification
Expression simplification allows us to make complex mathematical expressions more manageable. After using properties of square roots and absolute values, we simplified to find:
- \(|3c-3d| = 3|c-d|\).
- This is achieved by factoring out the constant 3, as it doesn’t affect the absolute value.
Other exercises in this chapter
Problem 108
Write the given expression without using absolute values. $$\frac{|u-v|}{|v-u|} \text { if } u \neq v, u \neq 0, v \neq 0$$
View solution Problem 109
Explain why the given statement is true for any numbers \(c\) and \(d .\) I Hint: Look at the properties of absolute value on page 10.1 $$\left|(c-d)^{2}\right|
View solution Problem 111
Express the given geometric statement about numbers on the number line algebraically, using absolute values. The distance from \(x\) to 5 is less than 4
View solution Problem 112
Express the given geometric statement about numbers on the number line algebraically, using absolute values. is more than 6 units from \(c\)
View solution