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Chapter 9: Problem 24

Solve the equation algebraically. Check the solutions graphically. $$ \frac{1}{3} x^{2}=3 $$

Short Answer

Expert verified
The algebraic solution to the equation \( \frac{1}{3} x^{2}=3 \) are \( x = \pm 3 \), these are also confirmed when checking graphically.

Step by step solution

01

Write the Equation in Standard Form

First transform the given equation into the standard form for quadratic equations. In this case, it means multiplying both sides of \( \frac{1}{3} x^{2}=3 \) by 3. Which outputs: \( x^{2}=9 \)
02

Take the Square Root of Both Sides

In order to find the roots or zeroes of the equation, the square root of both sides of the equation should be taken. So, \( sqrt{x^{2}} = \pm sqrt{9} \). This leads to: \( x = \pm 3 \)
03

Graph the Equation

Now, confirm the solutions graphically by plotting the original equation which will take a form of a parabola. Then check where it crosses the x-axis. As accurately as possible, based on the graph the intersection will be at -3 and 3, which confirms the solutions obtained algebraically.

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Key Concepts

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

Algebraic Solutions
Solving quadratic equations algebraically can seem tricky at first, but it's just a matter of following some simple steps. Every quadratic equation can be expressed in the standard form: \(ax^2 + bx + c = 0\). When solving equations of this form, we're essentially finding the values of \(x\) that make the equation true.

Here's how you can simplify and solve an equation using basic algebraic techniques:
  • Step 1: Express in Standard Form - Transform the original equation into the standard quadratic form. For instance, in the given problem \( \frac{1}{3}x^2 = 3 \), multiply both sides by 3 to eliminate the fraction, leading to \(x^2 = 9\).
  • Step 2: Solve for \(x\) - Once in standard form, solve for \(x\). Here, taking the square root of both sides yields \(x = \pm 3\). This step reveals that the solution consists of two possible values: 3 and -3.
By using algebraic methods, we can precisely identify the solutions, which are the zeros of the quadratic function.
Graphical Solutions
Graphical solutions offer a visual perspective that helps confirm algebraic results. When you graph a quadratic equation, you typically get a U-shaped curve known as a parabola.

To verify the solutions graphically:
  • Step 1: Plot the Parabola - Start by graphing the equation \(y = \frac{1}{3}x^2 - 3\). This is derived from rearranging the original equation \(\frac{1}{3}x^2 = 3\) into \(y = \frac{1}{3}x^2 - 3\).
  • Step 2: Identify the Intercepts - The points where the parabola intersects the x-axis represent the solutions to the equation. They confirm the values obtained algebraically (i.e., \(x = -3\) and \(x = 3\)). These are the x-intercepts of the graph.
Graphical solutions are useful as they provide a quick check and aid in visualizing how equations behave, showing the nature and number of solutions.
Square Root Method
The square root method is one of the simplest techniques to solve quadratic equations, useful when the equation is devoid of the linear term (i.e., \(bx\) in \(ax^2 + bx + c = 0\)).

To employ the square root method:
  • Step 1: To find the roots, if the equation is in the form \(x^2 = d\), simply take the square root of both sides. For example, when \( x^2 = 9 \), you find \(x\) by calculating the square root: \(x = \pm \sqrt{9}\).
  • Step 2: Simplify - In this case, simplify to find \(x = \pm 3\), which means both 3 and -3 satisfy the equation.
Using the square root method provides a direct route to solutions, making it an excellent choice for simple equations where no middle term is present.

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Most popular questions from this chapter

Solve the inequality and graph the solution. -3<-x<1

Sketch the graph of the function. Label the vertex. y=6 x^{2}-4 x-1

Sketch the graph of the function. Label the vertex. y=\frac{1}{2} x^{2}+2 x-1

LOGICAL REASONING Consider the equation \(a x^{2}+b x+c=0\) and use the quadratic formula to justify the statement. If \(b^{2}-4 a c\) is negative, then the equation has no real solution.

Solve the equation or write no solution. Write the solutions as integers if possible. Otherwise write them as radical expressions. $$x^{2}=-9$$
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