/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Solve the polynomial equation. I... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the polynomial equation. In Exercises \(7-14,\) find all solutions. In Exercises \(15-18,\) find only real solutions. Check your solutions. $$x^{4}-10 x^{2}=-21$$

Short Answer

Expert verified
The solutions are \(x = \pm\sqrt{7}\) and \(x = \pm\sqrt{3}\)

Step by step solution

01

Simplify the equation

Rearrange and simplify the equation to standard quadratic form. As \(x^4\) can be viewed as \((x^2)^2\), it is possible to consider the equation as a quadratic equation with variable \(x^2\). Therefore, rearrange the equation to get \((x^2)^2 - 10x^2 + 21 = 0\)
02

Solve the quadratic equation

Now solve the quadratic equation \(t^2 - 10t + 21 = 0\), where \(t = x^2\). The roots can be found using the quadratic formula \(t = (10 ± \sqrt{(-10)^2 - 4(1)(21)}) / 2(1)\). Solving this gives the roots \(t = 7\) and \(t = 3\)
03

Find actual solutions for x

Now that the solutions for \(t = x^2\) are found, convert them into possible solutions for \(x\) by taking square roots. From \(t = x^2\), we have \(x = \sqrt{t}\) and \(x = -\sqrt{t}\). Therefore, the solutions of the equation will be \(x = \pm\sqrt{7}\) and \(x = \pm\sqrt{3}\)
04

Check the solutions

Substitute the solutions \(x = \pm\sqrt{7}\) and \(x = \pm\sqrt{3}\) in the original equation to see if LHS equals to RHS. This will validate that the solutions are correct.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equations
Quadratic equations form an essential branch of algebra that is vital for understanding polynomial equations. Think of quadratic equations as expressing a relationship where an unknown variable is squared. A typical form looks like this:
  • \( ax^2 + bx + c = 0 \)
where \(a\), \(b\), and \(c\) are coefficients and \(a eq 0\). These equations represent parabolas when plotted on a graph.

In solving such equations, we often use techniques such as factoring, completing the square, or the quadratic formula. Each method can help pinpoint the points where the curve intersects the x-axis, known as the 'roots' of the equation. Quadratic equations can manipulate equations containing higher powers, such as a fourth power, as shown in the sample problem by redefining variables.

In the equation \( (x^2)^2 - 10x^2 + 21 = 0 \), we treat \( x^2 \) like a single bundle, making it easier to solve using the familiar quadratic process. This strategic complication reduction allows complex equations to be maneuvered and managed effectively.
Roots of Equations
The roots of an equation, especially in the context of quadratics, are essentially the solutions we find when setting the equation equal to zero. These roots tell us where the graph of the equation crosses the x-axis. For a quadratic equation, like \( ax^2 + bx + c = 0 \), roots can be found using:
  • Factoring the equation
  • Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
  • Completing the square, though less commonly used
Roots can be real or complex, depending on the equation's nature.

In the provided exercise, after transforming the original equation, we solved \( t^2 - 10t + 21 = 0 \) to find \( t = 7 \) and \( t = 3 \). These represent the squared forms of the final solutions. When viewed in the polynomial's original context, such roots were transformed back to the actual variable by taking the square roots. Including both positive and negative roots is critical as it represents the entry and exit points of the curve."
Real Solutions
Looking at real solutions is crucial for understanding which values satisfy the equation within the realm of real numbers. Real solutions are the x-values where the curve intersects the x-axis on a graph, visible and measurable in real life.
We determine if a quadratic has real solutions by checking the discriminant \( b^2 - 4ac \) from the quadratic formula. If it:
  • Is positive, the equation has two distinct real solutions.
  • Is zero, there is exactly one real solution, where the curve just touches the x-axis.
  • Is negative, there are no real solutions (solutions exist but are complex).
For example, in the solved equation, the discriminant was 16, a positive number, allowing the real solutions \( x = \pm \sqrt{7} \) and \( x = \pm \sqrt{3} \).

The interpretation of such real solutions helps in practical applications like physics and engineering, where only realistic, tangible values can apply.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Is it possible for a quadratic function to have the set of all real numbers as its range? Explain. (Hint: Examine the graph of a general quadratic function.)

When designing buildings, engineers must pay careful attention to how different factors affect the load a structure can bear. The following table gives the load in terms of the weight of concrete that can be borne when threaded rod anchors of various diameters are used to form joints. $$\begin{array}{cc} \text { Diameter (in.) } & \text { Load (1b) } \\ \hline 0.3750 & 2105 \\ 0.5000 & 3750 \\ 0.6250 & 5875 \\ 0.7500 & 8460 \\ 0.8750 & 11,500 \end{array}$$ (a) Examine the table and explain why the relationship between the diameter and the load is not linear. (b) The function $$f(x)=14,926 x^{2}+148 x-51$$ gives the load (in pounds of concrete) that can be borne when rod anchors of diameter \(x\) (in inches) are employed. Use this function to determine the load for an anchor with a diameter of 0.8 inch. (c) since the rods are drilled into the concrete, the manufacturer's specifications sheet gives the load in terms of the diameter of the drill bit. This diameter is always 0.125 inch larger than the diameter of the anchor. Write the function in part (b) in terms of the diameter of the drill bit. The loads for the drill bits will be the same as the loads for the corresponding anchors. (Hint: Examine the table of values and see if you can present the table in terms of the diameter of the drill bit.)

In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=|x| ; g(x)=\frac{x}{x-3}$$

Let \(f(x)=a x+b\) and \(g(x)=c x+d,\) where \(a, b, c,\) and \(d\) are constants. Show that \((f+g)(x)\) and \((f-g)(x)\) also represent linear functions.

In Exercises \(105-110\), find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) \(h \neq 0,\) for the given function \(f\). $$f(x)=-x^{2}+x$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.