Problem 60
Question
\(59-66\) . Find the solutions of the inequality by drawing appropriate graphs. State each answer rounded to two decimals. $$ 0.5 x^{2}+0.875 x \leq 0.25 $$
Step-by-Step Solution
Verified Answer
The solutions are the interval [-2, 0.25].
1Step 1: Express the Inequality
We start by expressing the given inequality: \[ 0.5 x^2 + 0.875 x \leq 0.25 \]Rearrange it to set it to zero: \[ 0.5 x^2 + 0.875 x - 0.25 \leq 0 \].
2Step 2: Multiply Through by 2
To eliminate decimals, multiply every term in the inequality by 2: \[ 2(0.5 x^2) + 2(0.875 x) - 2(0.25) \leq 0 \] which simplifies to:\[ x^2 + 1.75x - 0.5 \leq 0 \].
3Step 3: Find the Roots Using the Quadratic Formula
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the roots. Here, \( a = 1 \), \( b = 1.75 \), and \( c = -0.5 \).Calculate the discriminant:\[ \Delta = (1.75)^2 - 4(1)(-0.5) = 3.0625 + 2 = 5.0625 \].Find the square root of the discriminant:\[ \sqrt{5.0625} \approx 2.25 \].Calculate the roots:\[ x_1 = \frac{-1.75 + 2.25}{2} = 0.25 \] and \[ x_2 = \frac{-1.75 - 2.25}{2} = -2 \].
4Step 4: Identify the Interval Solution
With roots \( x = 0.25 \) and \( x = -2 \), we know the quadratic will be between these roots for \( x^2 + 1.75x - 0.5 \leq 0 \).This can be verified by checking a test point in the interval, such as \( x = 0 \), which gives: \[ 0^2 + 1.75(0) - 0.5 = -0.5 \leq 0 \].Thus, the solution interval is \([-2, 0.25]\).
5Step 5: Draw the Graph to Confirm
Draw a graph of the quadratic function \( y = x^2 + 1.75x - 0.5 \). Observe where it is below the x-axis. The roots \( x = -2 \) and \( x = 0.25 \) should be clear, confirming the interval \([-2, 0.25]\) as the solution.
Key Concepts
Quadratic FormulaDiscriminantGraphical Solution of Inequalities
Quadratic Formula
The quadratic formula is a powerful tool used to find the solutions of quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). This formula helps us to find the roots (or solutions) of these equations by plugging the coefficients \( a \), \( b \), and \( c \) into the formula:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- \( -b \): This term changes the sign of \( b \) and is part of finding the coordinate for the axis of symmetry of the parabola.
- \( \sqrt{b^2 - 4ac} \): This is known as the discriminant, and it tells us about the nature and number of the roots.
- \( 2a \): This term in the denominator divides the result into two potential roots due to the \( \pm \) part of the formula, giving us two solutions.
Discriminant
The discriminant is a crucial part of the quadratic formula and it provides valuable insight into the nature of the roots of a quadratic equation. It is represented by the expression \( b^2 - 4ac \). Based on its value, we can determine:
Hence, understanding the discriminant helps in interpreting the solution set without necessarily graphing the equation first.
- If \( b^2 - 4ac > 0 \): The quadratic equation has two distinct real roots.
- If \( b^2 - 4ac = 0 \): There is exactly one real root, also known as a repeated or double root.
- If \( b^2 - 4ac < 0 \): The equation has two complex roots, which means they do not intersect the x-axis at any point.
Hence, understanding the discriminant helps in interpreting the solution set without necessarily graphing the equation first.
Graphical Solution of Inequalities
Graphically solving quadratic inequalities involves sketching the graph of the corresponding quadratic equation and identifying regions where the graph is below or above the x-axis, depending on the inequality sign. Here's how you can do it:
- First, transform the inequality to an equation by setting it as equals \( (ax^2 + bx + c = 0) \) to find the roots.
- Plot these roots on a coordinate grid, which shows where the parabola intersects the x-axis.
- The parabola's direction can be determined from the sign of \( a \). Here, \( a = 1 \), indicating an upward opening parabola.
- Check the portions of the parabola relative to the x-axis. For \( ax^2 + bx + c \leq 0 \), identify the interval where the parabola lies below the x-axis.
Other exercises in this chapter
Problem 59
\(55-62\) . Find an equation of the circle that satisfies the given conditions. Endpoints of a diameter are \(P(-1,1)\) and \(Q(5,9)\)
View solution Problem 60
Use slopes to determine whether the given points are collinear (lie on a line). $$ \begin{array}{l}{\text { (a) }(1,1),(3,9),(6,21)} \\ {\text { (b) }(-1,3),(1,
View solution Problem 60
\(55-62\) . Find an equation of the circle that satisfies the given conditions. Endpoints of a diameter are \(P(-1,3)\) and \(Q(7,-5)\)
View solution Problem 61
Find an equation of the perpendicular bisector of the line segment joining the points \(A(1,4)\) and \(B(7,-2)\)
View solution