Problem 61 - Algebra Form and Function

Question

Which of the following equations have the same solutions as the equation \(9 x^{2}=81 ?\) (a) \(3 x=9\) (b) \(9 x=\pm 9\) (c) \(3 x=\pm 9\) (d) \(x^{2}=9\)

Step-by-Step Solution

Answer

Answer: Equations (c) \(3x = \pm 9\) and (d) \(x^2 = 9\) have the same solutions as the main equation \(9x^2=81\).

Step 1: Solve the main equation

First, we will solve the main equation \(9x^2 = 81\) to find its solutions. Divide both sides of the equation by 9: \(x^2 = \frac{81}{9}\) \(x^2 = 9\) Now, find the square root of 9: \(x = \pm \sqrt{9}\) \(x = \pm 3\) So the solutions for the main equation are \(\pm 3\).

Step 2: Solve equation (a)

Now, we will solve equation (a) \(3x = 9\). To do this, divide both sides by 3: \(x = \frac{9}{3}\) \(x = 3\) The only solution for equation (a) is 3, which is not equal to the main equation's solutions.

Step 3: Solve equation (b)

For equation (b) \(9x = \pm 9\), we will divide both sides by 9: \(x = \frac{\pm 9}{9}\) \(x = \pm 1\) The solutions for equation (b) are \(\pm 1\), which are not equal to the main equation's solutions.

Step 4: Solve equation (c)

For equation (c) \(3x = \pm 9\), we will divide both sides by 3: \(x = \frac{\pm 9}{3}\) \(x = \pm 3\) The solutions for equation (c) are \(\pm 3\), which are equal to the main equation's solutions.

Step 5: Solve equation (d)

Finally, we will solve equation (d) \(x^2 = 9\): \(x = \pm \sqrt{9}\) \(x = \pm 3\) The solutions for equation (d) are \(\pm 3\), which are equal to the main equation's solutions.

Step 6: Conclusion

Based on the solutions of the given equations, only equations (c) \(3x = \pm 9\) and (d) \(x^2 = 9\) have the same solutions as the main equation \(9 x^{2} = 81\).

Key Concepts

Equivalent Equations

Understanding equivalent equations is key in solving algebraic problems efficiently. Equivalent equations are different expressions that have the same solution set. For instance, if we start with the equation \(9x^2 = 81\), we can transform it by dividing every term by 9 to get \(x^2 = 9\). These are equivalent because they yield the same solutions.

Translating an equation into an equivalent form often involves simplification or customization to ease solving. Here is what you can usually do:

• Divide or multiply all terms by the same number, as long as it isn't zero.
• Add or subtract the same value from both sides.
• Use algebraic identities, like factoring.

Recognizing equivalent equations means you can identify potential shortcuts and instinctively know that different-looking equations might just be different paths to the same solution.

Square Roots

Square roots are used to reverse the operation of squaring a number. When solving equations like \(x^2 = 9\), finding the square root is vital in determining possible solutions. Numbers have two square roots: one positive and one negative, because squaring any positive or negative number returns a positive result.

For instance, the square roots of 9 are 3 and -3 since both \(3 \times 3 = 9\) and \(-3 \times -3 = 9\). When solving quadratic equations, always remember to consider both roots in order to capture all possible solutions.

• Square roots are represented using the symbol \(\sqrt{}\).
• To solve \(x^2 = a\), compute \(x = \pm \sqrt{a}\).

Square roots often lead to two solutions, which is why the solutions to the given equation \(9x^2 = 81\) include \(\pm 3\). Forgetting the negative root can result in missing solutions.

Solution Sets

Solution sets represent all possible values that satisfy an equation. In quadratic equations, recognizing all potential solutions is crucial. Take \(9x^2 = 81\) for instance. Solving it gives us \(x = \pm 3\), meaning the solution set consists of the numbers \(-3\) and \(3\).

Knowing how to determine and verify solution sets helps in confirming the accuracy of algebraic manipulations. For example:

• A solution set verifies the validity of equivalent equations, as seen with \(x^2 = 9\).
• It helps cross-verify results from multiple equations, such as matching the main question with options (c) and (d).
• It provides clear insight into the potential scenarios for the equation's outcome.

Always check that the solution set contains all possible answers by substituting them back into the original equation. This process ensures you have found every viable option.