help solve

Problem 1:

 

We can complete the square to minimize the given expression

 

Steps to solve:1. Complete the square

 

First, we recognize that the expression is already in the form of a perfect square trinomial up to a constant factor. This means we can complete the square by taking half of the coefficient of our x2 term, squaring it, and adding it to both sides of the equation.

 

In this case, the coefficient of our x2 term is −2, so half of it would be −1, and squaring it gives us 1. Because we're adding the constant term inside the parentheses on the right where it's being multiplied by 1 (the coefficient of x2 ), we need to add −1 (the opposite of half the coefficient of our x2 term squared) outside the parentheses on the left to make sure we're adding the same thing to both sides.

 

x4−2x2−1=x4−2x2−1

 

(x2)2−2(x2)+1=(x2−1)2

 

2. Minimum value:

 

Now that the expression is a squared term, we know its minimum value is 0, which occurs when x2=1 or x=±1.

 

Answer: The minimum value of the expression is x2(x2−2) is 0,​ which occurs when x=±1.

Problem 2:

 

To find the largest possible value of a + 2b + 3c, we want to maximize the values of a, b, and c.

Since a, b, and c can be the numbers 1, 2, 3 in some order, let's consider the case where a = 3, b = 2, and c = 1:

a + 2b + 3c = 3 + 2(2) + 3(1) = 3 + 4 + 3 = 10

Therefore, the largest possible value of a + 2b + 3c is 10.

Problem 3:

 

To find the integers n that satisfy 64≤n2≤100, we can take the square root of both sides of the inequality. However, since the square root function is not one-to-one (it has two outputs for each positive input), we need to be careful.

 

Taking the square root of both sides gives:

 

8≤∣n∣≤10

 

Since we only care about positive integer solutions for n, we can simply remove the absolute value signs and consider 8≤n≤10.

 

Now, we can count the number of integers in this range, inclusive. There are 10−8+1=3​ integers that satisfy the inequality: 8, 9, and 10.