Square of Difference: Identify the greater value

To solve this problem, we'll follow these steps:

• Step 1: Expand both expressions.
• Step 2: Subtract one expression from the other to determine the inequality.
• Step 3: Analyze and conclude based on the comparison.

Now, let's work through each step:
Step 1: Expand $(x-4)^2$:

$(x-4)^2 = (x-4)(x-4) = x^2 - 2 \cdot 4 \cdot x + 16 = x^2 - 8x + 16$

Step 2: Set up the inequality $(x-4)^2 < x^2 + 16$ and substitute the expanded form:

$x^2 - 8x + 16 < x^2 + 16$

Step 3: Simplify the inequality by subtracting $x^2$ and $16$ from both sides:

$x^2 - 8x + 16 - x^2 - 16 < 0$

$-8x < 0$

Solving $-8x < 0$ gives:

$x > 0$

This inequality holds true for all $x > 0$.

Therefore, the inequality $(x-4)^2 < x^2 + 16$ is correct for $x > 0$.

Thus, the correct symbol to fill in the blank is $\lt$.

To solve this problem, we'll follow these steps:

• Step 1: Simplify the expression $(\sqrt{40} - \sqrt{10})^2$
• Step 2: Calculate the square root values for $\sqrt{40}$ and $\sqrt{10}$
• Step 3: Compare the simplified value to 30

Now, let's work through each step:
Step 1: Simplify $(\sqrt{40} - \sqrt{10})^2$ using the square of a difference formula:

$(\sqrt{40} - \sqrt{10})^2 = (\sqrt{40})^2 - 2 \cdot \sqrt{40} \cdot \sqrt{10} + (\sqrt{10})^2$

$(\sqrt{40})^2 = 40$ and $(\sqrt{10})^2 = 10$, so:

$(\sqrt{40} - \sqrt{10})^2 = 40 - 2 \cdot \sqrt{40} \cdot \sqrt{10} + 10$

Step 2: Calculate each term:
$\sqrt{40} \approx 6.324$ and $\sqrt{10} \approx 3.162$.

The expression becomes:

$40 - 2 \cdot 6.324 \cdot 3.162 + 10 = 50 - 40 \approx 10$

Step 3: Compare $10$ to $30$. Since $10 < 30$, the correct relation is:

Therefore, the solution to the problem is the comparison $(\sqrt{40} - \sqrt{10})^2 < 30$.

Thus, the sign is $\textbf{<}$.

The solution to the comparison problem is $=$.

To solve this problem, we'll follow these steps:

• Step 1: Simplify the expression $(a-\sqrt{20})^2$.
• Step 2: Simplify the expression $a(a+20) + 20$.
• Step 3: Compare the results of the two expressions.

Now, let's work through each step:

Step 1: We start with the expression $(a-\sqrt{20})^2$. Using the formula for the square of a difference, we get:

$(a-\sqrt{20})^2 = a^2 - 2a\sqrt{20} + 20$.

Step 2: Now we consider the expression $a(a+20) + 20$. Expanding this, we have:

$a(a+20) + 20 = a^2 + 20a + 20$.

Step 3: Now, we compare the two simplified expressions:

$a^2 - 2a\sqrt{20} + 20$ and $a^2 + 20a + 20$.

Both sides share an $a^2 + 20$, so we compare the remaining terms:

$-2a\sqrt{20}$ and $20a$.

Rewriting these as inequalities, since $0 < a$ and $-2\sqrt{20} \approx -8.944$, which is smaller than $20$. This gives:

$-2a\sqrt{20} < 20a$.

Thus, $(a-\sqrt{20})^2 < a(a+20) + 20$.

Therefore, the correct comparison sign is < .

To solve this problem, we need to compare two expressions:

• Expression 1: $(3b-4)^2$, which can be expanded as:

$(3b-4)^2 = 9b^2 - 24b + 16$

• Expression 2: $9b\left(b+\frac{1}{b}\right) + 7$, which expands to:

$9b \cdot b + 9b \cdot \frac{1}{b} + 7 = 9b^2 + 9 + 7 = 9b^2 + 16$

Now, we compare the simplified expressions:

• Expression 1: $9b^2 - 24b + 16$
• Expression 2: $9b^2 + 16$

The only difference between these expressions is the $-24b$ term in Expression 1, which makes it smaller since $-24b$ is negative for $b > 0$.

Therefore, $(3b-4)^2$ is less than $9b(b+\frac{1}{b})+7$. The correct inequality sign is < .

The task is to determine the inequality $a(a-b)^2$ compared to $a(a+b)^2 - ab^2$ given $0 < a, b$. Here's how to solve it:

First, let's expand and simplify each expression:

Starting with $a(a-b)^2$, expand as follows:

• $(a-b)^2 = a^2 - 2ab + b^2$.
• Multiplying by $a$ gives: $a(a^2 - 2ab + b^2) = a^3 - 2a^2b + ab^2$.

Next, expand $a(a+b)^2-ab^2$:

• $(a+b)^2 = a^2 + 2ab + b^2$.
• Multiplying by $a$ gives: $a(a^2 + 2ab + b^2) = a^3 + 2a^2b + ab^2$.
• Subtracting $ab^2$ yields: $a^3 + 2a^2b + ab^2 - ab^2 = a^3 + 2a^2b$.

Now compare the two expressions:

• $a(a-b)^2 = a^3 - 2a^2b + ab^2$.
• $a(a+b)^2 - ab^2 = a^3 + 2a^2b$.
• The inequality $a(a-b)^2 ? a(a+b)^2 - ab^2$ is simplified to:\)
• $a^3 - 2a^2b + ab^2 < a^3 + 2a^2b$.
• Cancel $a^3$ from both sides: $-2a^2b + ab^2 < 2a^2b$.
• Bring like terms together: $-4a^2b + ab^2 < 0$.
• Factor $b$: $b(-4a^2 + ab) < 0$.

Given $b > 0$, divide through by $b$:

$-4a^2 + ab < 0$, or equivalently $-4a^2 < -ab$, hence $4a > b$.

Therefore, $a > \frac{b}{4}$.

The inequality holds as $a(a-b)^2 < a(a+b)^2 - ab^2$ when $a > \frac{b}{4}$.

Therefore, the correct choice is:

$<$ When $a > \frac{b}{4}$