Problem 30
Question
Solve each equation analyrically for all complex solutions, giving exact forms in your solution set. Then graph the left side of the equation as \(Y_{1}\) in the suggested viewing window and, using the capabilities of vour calculater, suppert the real solutions. $$\begin{aligned}&-4 x^{3}-x^{2}+4 x=0\\\&[-4,4] \text { by }[-10,10]\end{aligned}$$
Step-by-Step Solution
Verified Answer
Solutions: \(x = 0, \frac{1+\sqrt{65}}{-8}, \frac{1-\sqrt{65}}{-8}\); real solutions confirmed by graph.
1Step 1: Factor the Equation
First, factor the left side of the equation \(-4x^3 - x^2 + 4x = 0\).Start by factoring out the common factor of \(x\):\[x(-4x^2 - x + 4) = 0\]Now, solve \(-4x^2 - x + 4 = 0\) separately.
2Step 2: Apply the Quadratic Formula
To solve the quadratic equation \(-4x^2 - x + 4 = 0\), we use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = -4\), \(b = -1\), and \(c = 4\). Plug these into the formula:\[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(-4)(4)}}{2(-4)}\]Simplifying further gives:\[x = \frac{1 \pm \sqrt{1 + 64}}{-8}\]\[x = \frac{1 \pm \sqrt{65}}{-8}\]
3Step 3: Solve the Linear Factor
The factored equation from Step 1 gives us the linear factor \(x = 0\). Therefore, the equation has at least one solution where \(x = 0\).
4Step 4: Identify All Solutions
The complete set of solutions for the equation includes the linear solution from Step 3 and the solutions from the quadratic solution in Step 2:\[x = 0\]\[x = \frac{1 + \sqrt{65}}{-8}\]\[x = \frac{1 - \sqrt{65}}{-8}\]Thus, the exact complex solutions to the equation are \(x = 0, \frac{1 + \sqrt{65}}{-8}, \) and \( \frac{1 - \sqrt{65}}{-8}\).
5Step 5: Graph and Confirm Real Solutions
Using a calculator, graph \(-4x^3 - x^2 + 4x\) over the viewing window \([-4, 4]\) by \([-10, 10]\). Observe where the graph crosses the x-axis to confirm real solutions.The real solutions are confirmed to be \(x = 0\) and the approximate values of the quadratic roots. Verify the solutions by checking the graph.
Key Concepts
Factoring EquationsQuadratic FormulaGraphing EquationsReal Solutions Verification
Factoring Equations
Factoring equations is a key skill in algebra that helps to break down complex expressions into simpler components. In the context of solving polynomial equations, factoring allows us to express the equation as a product of simpler expressions, which can then be solved independently.
In our exercise, we tackled the polynomial \(-4x^3 - x^2 + 4x = 0\). The first step was to identify a common factor in each term, which is \(x\). Factoring out \(x\) gives us:
In our exercise, we tackled the polynomial \(-4x^3 - x^2 + 4x = 0\). The first step was to identify a common factor in each term, which is \(x\). Factoring out \(x\) gives us:
- \(x(-4x^2 - x + 4) = 0\)
Quadratic Formula
The quadratic formula is an essential tool for finding solutions to quadratic equations. These are equations in the form \(ax^2 + bx + c = 0\). The formula is given by:
- \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
- \(x = \frac{1 \pm \sqrt{65}}{-8}\)
Graphing Equations
Graphing equations allows us to visually interpret the solutions and behavior of the function. Using a graphing calculator, plotting the equation helps to confirm the roots and understand their positions relative to the x-axis.
In this exercise, we graphed \(-4x^3 - x^2 + 4x\) in the window \([-4, 4]\) by \([-10, 10]\). The graph provided a clear picture of where the polynomial crosses the x-axis, showing us the real intersections:
The crossing points on the x-axis at \(x = 0\) and around the approximate roots given by the quadratic formula show the real solutions. Graphs are an invaluable tool for supporting analytic solutions, offering a means to verify and visualize our results.
In this exercise, we graphed \(-4x^3 - x^2 + 4x\) in the window \([-4, 4]\) by \([-10, 10]\). The graph provided a clear picture of where the polynomial crosses the x-axis, showing us the real intersections:
The crossing points on the x-axis at \(x = 0\) and around the approximate roots given by the quadratic formula show the real solutions. Graphs are an invaluable tool for supporting analytic solutions, offering a means to verify and visualize our results.
Real Solutions Verification
Verification of real solutions is a critical step in solving equations to ensure accuracy. Once we have the potential solutions, both from factoring and the quadratic formula, verifying them against the graph is crucial.
In this problem, we identified potential solutions \(x = 0\) and the roots from the quadratic formula as real. By graphing the equation, we checked where the graph crosses the x-axis:
These real intersections confirmed our solutions, showing \(x = 0\) visibly. The approximate intersection points align with the calculation results from the quadratic, ensuring that our analytic methods were correct. Verifying solutions using graphs provides a practical tool to double-check work and understand the nature of the roots.
In this problem, we identified potential solutions \(x = 0\) and the roots from the quadratic formula as real. By graphing the equation, we checked where the graph crosses the x-axis:
These real intersections confirmed our solutions, showing \(x = 0\) visibly. The approximate intersection points align with the calculation results from the quadratic, ensuring that our analytic methods were correct. Verifying solutions using graphs provides a practical tool to double-check work and understand the nature of the roots.
Other exercises in this chapter
Problem 30
Find each quotient when \(P(x)\) is divided by the specified binomial. $$P(x)=-3 x^{3}-x-5 ; \quad x+1$$
View solution Problem 30
Find a polynomial function \(P(x)\) having leading coefficient 1, least possible degree, real coefficients. and the given zeros. 5 (multiplicity 2 ) and \(-2 i\
View solution Problem 31
Sketch by hand the graph of each function. (You may wish to support your answer with a calculator graph.) $$\begin{aligned} P(x) &=2 x^{3}-5 x^{2}-x+6 \\ &=(x+1
View solution Problem 31
Find each quotient when \(P(x)\) is divided by the specified binomial. $$P(x)=x^{5}-1 ; x-1$$
View solution