Algebraic Inequality Solution: -2x^2 + 146 <= 2x + 2

<= 2x + 2$

Let's dive into solving the inequality $-2x^2 + 146 <= 2x + 2$ algebraically. Inequalities, much like equations, involve finding the values of a variable that make a statement true. However, instead of a single solution (or a few discrete solutions), inequalities often yield a range of values. When we're dealing with a quadratic inequality like this one, we're essentially looking for the intervals on the number line where the quadratic expression is less than or equal to another expression. The key to tackling this is to first transform it into a standard quadratic inequality form, usually by moving all terms to one side so that we have zero on the other. This makes it easier to analyze the behavior of the quadratic function. Once we have it in that form, we'll find the roots of the corresponding quadratic equation. These roots act as critical points, dividing the number line into distinct intervals. We then test a value from each interval to see if it satisfies the original inequality. This systematic approach ensures we don't miss any part of the solution.

Rearranging the Inequality

Our first step in solving $-2x^2 + 146 <= 2x + 2$ is to consolidate all terms onto one side to get a standard quadratic inequality. It's often convenient to have the $x^2$ term with a positive coefficient, but for inequalities, it's not strictly necessary as long as we are careful with our steps. Let's move all terms to the right side to make the $x^2$ coefficient positive. Subtract 146 from both sides: $-2x^2 <= 2x + 2 - 146$, which simplifies to $-2x^2 <= 2x - 144$. Now, subtract $2x$ from both sides: $-2x^2 - 2x <= -144$. Finally, add 144 to both sides: $-2x^2 - 2x + 144 <= 0$. We now have our inequality in the form $ax^2 + bx + c <= 0$. For ease of calculation and to avoid potential sign errors when factoring or using the quadratic formula, we can divide the entire inequality by -2. Crucially, when we divide or multiply an inequality by a negative number, we must reverse the direction of the inequality sign. So, dividing by -2, we get $x^2 + x - 72

= 0$. This is the form we'll work with moving forward. This rearranged form, $x^2 + x - 72 = 0$, is much more manageable and directly comparable to our analysis of the roots.

Finding the Critical Points (Roots)

Now that we have the inequality in the form $x^2 + x - 72

= 0$, we need to find the roots of the corresponding quadratic equation, which is $x^2 + x - 72 = 0$. These roots are the points where the quadratic function $f(x) = x^2 + x - 72$ crosses the x-axis. They are critical because they divide the number line into intervals where the function's value is either positive or negative. We can find these roots using factoring, completing the square, or the quadratic formula. Factoring is often the quickest method if it's possible. We're looking for two numbers that multiply to -72 and add up to 1 (the coefficient of the x term). Let's consider pairs of factors of 72: (1, 72), (2, 36), (3, 24), (4, 18), (6, 12), (8, 9). To get a product of -72, one factor must be positive and the other negative. To get a sum of +1, the numbers must be close together, and the larger number must be positive. The pair (8, 9) fits this. If we have -8 and +9, their product is -72 and their sum is +1. Therefore, we can factor the quadratic as $(x - 8)(x + 9) = 0$. Setting each factor to zero, we get $x - 8 = 0$ (which gives $x = 8$) and $x + 9 = 0$ (which gives $x = -9$). These are our critical points: $x = -9$ and $x = 8$. These values are where the expression $x^2 + x - 72$ equals zero. They are the boundaries that will help us determine the intervals where the expression is greater than or equal to zero.

Analyzing the Intervals

With our critical points $x = -9$ and $x = 8$, we can now divide the number line into three distinct intervals: $(- ext{infinity}, -9)$, $(-9, 8)$, and $(8, ext{infinity})$. Our goal is to determine in which of these intervals the inequality $x^2 + x - 72

= 0$ holds true. Since the quadratic expression $x^2 + x - 72$ represents a parabola opening upwards (because the coefficient of $x^2$ is positive, which is 1), we know that the parabola will be above the x-axis (positive) outside its roots and below the x-axis (negative) between its roots. However, it's always a good practice to test a value from each interval to confirm.

• Interval 1: $(- ext{infinity}, -9)$. Let's pick a test value, say $x = -10$. Plugging this into $x^2 + x - 72$: $(-10)^2 + (-10) - 72 = 100 - 10 - 72 = 90 - 72 = 18$. Since $18

= 0$, this interval satisfies the inequality.

• Interval 2: $(-9, 8)$. Let's pick a test value, say $x = 0$. Plugging this into $x^2 + x - 72$: $(0)^2 + (0) - 72 = -72$. Since $-72$ is not $>= 0$, this interval does not satisfy the inequality.
• Interval 3: $(8, ext{infinity})$. Let's pick a test value, say $x = 10$. Plugging this into $x^2 + x - 72$: $(10)^2 + (10) - 72 = 100 + 10 - 72 = 110 - 72 = 38$. Since $38

= 0$, this interval satisfies the inequality.

Also, remember that the inequality is $>= 0$, meaning the points where the expression equals zero are also part of our solution. So, our critical points $x = -9$ and $x = 8$ are included in the solution set.

The Solution Set

Based on our analysis of the intervals, the inequality $x^2 + x - 72

= 0$ is satisfied for $x$ values in the interval $(- ext{infinity}, -9]$ and $[8, ext{infinity})$. The square brackets indicate that the endpoints -9 and 8 are included in the solution set because the inequality is "greater than or equal to." Therefore, the solution to the original inequality $-2x^2 + 146 <= 2x + 2$ is all real numbers $x$ such that $x <= -9$ or $x = 8$. In interval notation, this is represented as $(- ext{infinity}, -9] ext{ U } [8, ext{infinity})$. This means any number less than or equal to -9, or any number greater than or equal to 8, will make the original statement true. It's always a good idea to double-check a value from each part of the solution and one from the excluded region to ensure accuracy. For example, let's check $x=-10$ (should work): $-2(-10)^2 + 146 = -2(100) + 146 = -200 + 146 = -54$. And $2(-10) + 2 = -20 + 2 = -18$. Is $-54 <= -18$? Yes, it is. Now let's check $x=10$ (should work): $-2(10)^2 + 146 = -2(100) + 146 = -200 + 146 = -54$. And $2(10) + 2 = 20 + 2 = 22$. Is $-54 <= 22$? Yes, it is. Let's check a value in the excluded region, say $x=0$: $-2(0)^2 + 146 = 146$. And $2(0) + 2 = 2$. Is $146 <= 2$? No, it is not. This confirms our solution set is correct.

For more in-depth understanding of inequalities and quadratic functions, you can explore resources like Khan Academy or Paul's Online Math Notes.