Solving 2x^2+6x-13 > -x+2: A Non-linear Inequality

Let's explore a quadratic inequality and understand what we're trying to solve.Here's our inequality. We need to find all x values that make this statement true.The left side contains a quadratic expression with an x squared term.While the right side is a simpler linear expression with just x to the first power.Let's break down the quadratic expression. It has three parts: a quadratic term, a linear term, and a constant term.The inequality symbol tells us that the left side must be greater than the right side.We need to find all x values on the number line that make this inequality true.Before we solve this, let's note some important points about quadratic inequalities.Now that we understand what we're looking for, let's learn how to solve this inequality.To solve this inequality, we need to get all terms on one side of the inequality sign.First, we'll move negative x and positive two from the right side to the left side. Remember, when moving terms across the inequality sign, we change their signs.Now we can combine like terms. We have six x plus x, which gives us seven x.We also combine the constant terms: negative thirteen minus two equals negative fifteen.This gives us our quadratic expression in standard form: two x squared plus seven x minus fifteen is greater than zero.This is the standard form we need for solving the quadratic inequality. Notice all terms are on the left side, and we have zero on the right.In this form, we can clearly see the coefficients of our quadratic expression, which we'll need for the next step.Now that we have our inequality in standard form, we can move on to finding its solution.Now that we have our quadratic expression in standard form, let's identify each coefficient.The standard form of a quadratic expression is a x squared plus b x plus c.First, let's identify a, the coefficient of x squared. In our expression, a equals 2.Next, b is the coefficient of x. Here, b equals 7.Finally, c is our constant term, which is negative 15.We'll need these coefficients when we use the quadratic formula to find the roots of our equation.In the next step, we'll substitute a equals 2, b equals 7, and c equals negative 15 into this formula.Now that we've identified our coefficients, we're ready to solve the quadratic equation.Now we'll apply the quadratic formula to find the x-intercepts of our quadratic expression.Let's identify our values: a is 2, the coefficient of x squared; b is 7, the coefficient of x; and c is negative 15, our constant term.Let's substitute these values into the quadratic formula.First, we'll simplify b squared, which is seven squared, giving us forty-nine.Next, we multiply negative four times a times c, which is negative four times two times negative fifteen.This gives us negative four times negative thirty, which is positive one hundred twenty.Under the square root, we now have forty-nine plus one hundred twenty, which simplifies to one hundred sixty-nine.The square root of one hundred sixty-nine is thirteen, giving us our final form of negative seven plus or minus thirteen, all over four.This expression will give us the two x-intercepts of our quadratic function when we evaluate the plus and minus cases separately.Now that we have simplified the square root, we can calculate our critical points.Let's calculate the positive solution first.And now the negative solution.Let's visualize these points on a coordinate plane.These critical points, negative five and three halves, are where our quadratic expression equals zero.The parabola crosses the x-axis at these two points.These x-intercepts are crucial for determining where our quadratic inequality changes sign.In the next section, we'll analyze what happens between and outside these points.Now that we have our critical points, let's place them on a number line.The critical points are negative five and three halves.Since our parabola opens upward, the expression is positive outside these points.Let's test a point in each interval to verify our solution.For x equals negative five point five, in our left interval.For x equals zero, in our middle interval.And for x equals two, in our right interval.These calculations confirm that our expression is positive when x is less than negative five or greater than three halves.To verify our solution, we'll test points in each interval of our number line.We'll test three strategic points: negative six in the left interval, zero in the middle interval, and two in the right interval.Let's start with x equals negative six. Substituting into our expression two x squared plus seven x minus fifteen.Next, we'll test x equals zero in the middle interval.Finally, let's test x equals two in the right interval.Our calculations confirm that the expression is positive when x is less than negative five or greater than three halves.Now that we've confirmed our intervals, we can write our final solution.Now that we've found our critical points, let's write our solution in proper interval notation.Our solution includes all values less than negative fiveAnd all values greater than three halvesIn interval notation, we express this as the union of two intervals: from negative infinity to negative five, open parentheses, union with three halves to infinity, also with open parentheses.Let's break down this notation: The parentheses mean we don't include the endpoints, infinity symbols represent all numbers extending in that direction, and the union symbol means 'or'.Remember, this means our original inequality is satisfied when x is either less than negative five or greater than three halves.To verify our solution, let's test several points in our original inequality.We'll create a table to systematically check different x values.Let's start with x equals negative six, which is in our first solution interval.At x equals negative five, our boundary point, let's verify the inequality.Testing x equals zero, which is in the excluded middle interval.At x equals three halves, our other boundary point.Finally, let's test x equals two, which is in our second solution interval.Our verification confirms that points in the solution intervals satisfy the inequality, while points outside do not.These test points verify that our solution is correct.