Compare (a-√20)² and a(a+20)+20: Algebraic Inequality Challenge

Question

Since 0 < a Fill in the correct sign

(a20)2?a(a+20)+20 (a-\sqrt{20})^2?a(a+20)+20

Video Solution

Solution Steps

00:00 Complete the appropriate sign
00:04 We will use shortened multiplication formulas to open the parentheses
00:24 We will properly open parentheses, multiply by each factor
00:47 We will reduce what we can
01:08 Negative is of course less than positive
01:13 And this is the solution to the question

Step-by-Step Solution

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

  • Step 1: Simplify the expression (a20)2 (a-\sqrt{20})^2 .
  • Step 2: Simplify the expression a(a+20)+20 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 (a20)2 (a-\sqrt{20})^2 . Using the formula for the square of a difference, we get:

(a20)2=a22a20+20 (a-\sqrt{20})^2 = a^2 - 2a\sqrt{20} + 20 .

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

a(a+20)+20=a2+20a+20 a(a+20) + 20 = a^2 + 20a + 20 .

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

a22a20+20 a^2 - 2a\sqrt{20} + 20 and a2+20a+20 a^2 + 20a + 20 .

Both sides share an a2+20 a^2 + 20 , so we compare the remaining terms:

2a20-2a\sqrt{20} and 20a20a.

Rewriting these as inequalities, since 0<a 0 < a and 2208.944-2\sqrt{20} \approx -8.944 , which is smaller than 20 20 . This gives:

2a20<20a -2a\sqrt{20} < 20a .

Thus, (a20)2<a(a+20)+20 (a-\sqrt{20})^2 < a(a+20) + 20 .

Therefore, the correct comparison sign is < .

Answer

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