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Chapter 3: Problem 60

Find all real number solutions for each equation. $$4 y^{2}=25$$

Short Answer

Expert verified
The solutions are \(y = \frac{5}{2}\) and \(y = -\frac{5}{2}\).

Step by step solution

01

Isolate the variable term

We start with the equation \(4y^{2} = 25\). Our goal is to solve for \(y\). To begin, divide both sides of the equation by 4 to isolate \(y^2\):\[y^{2} = \frac{25}{4}\]
02

Apply the square root

To solve for \(y\), we need to take the square root of both sides of the equation \(y^{2} = \frac{25}{4}\). Remember that taking the square root of both sides introduces both positive and negative solutions:\[y = \pm \sqrt{\frac{25}{4}}\]
03

Simplify the square root

Now simplify the square root. Since \(\sqrt{\frac{25}{4}}\) can be broken down to \(\frac{\sqrt{25}}{\sqrt{4}}\):\[y = \pm \frac{5}{2}\]The solutions are \(y = \frac{5}{2}\) and \(y = -\frac{5}{2}\).

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Key Concepts

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

Real Number Solutions
When we're asked to find real number solutions for an equation, it means we need the solutions to be within the real numbers. Real numbers include all the numbers on the number line, like 0, positive numbers, negative numbers, and all the fractions or decimals in between. These do not include imaginary numbers. Real number solutions allow us to find values that satisfy an equation under normal arithmetic operations.
  • Any equation, like the one we solved, \(4y^2 = 25\), may have zero, one, or multiple real number solutions depending on the circumstances.
  • The equation \(y^2 = \frac{25}{4}\) ultimately provides two real number solutions for \(y\): \(y = \frac{5}{2}\) and \(y = -\frac{5}{2}\).
The square root step is critical in revealing all possible real solutions as we consider both the positive and negative roots.
Square Roots
The square root operation helps us determine which numbers, when squared (multiplied by themselves), result in a given number. Here, finding the square root is essential for solving quadratic equations.
When working with the square root:
  • Recall that every positive real number has two square roots: a positive and a negative one. For example, both \(5\) and \(-5\) are square roots of \(25\) because \(5^2 = 25\) and \((-5)^2 = 25\).
  • In the equation \(y^2 = \frac{25}{4}\), we apply the square root to get \(y = \pm \frac{5}{2}\).
  • This indicates that \(y\) could either be \(\frac{5}{2}\) or \(-\frac{5}{2}\).
Be mindful that taking a square root can introduce both positive and negative solutions, which must be considered in the context of real numbers.
Isolating Variables
Isolating the variable involves rearranging an equation to express the variable of interest on one side of the equation by itself. This is often the first step when solving equations.
  • The goal is to manipulate the equation to make the term with the variable alone by adding, subtracting, multiplying, or dividing as needed.
  • In the given problem, we started with \(4y^2 = 25\) and divided both sides by 4 to isolate \(y^2\), leading to \(y^2 = \frac{25}{4}\).
By isolating \(y^2\) first, we set up an easier path to solve for \(y\) using square roots. This method of decomposition by isolating variables simplifies the problem and paves the way to arrive at valid solutions.

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Most popular questions from this chapter

Factor each trinomial and assume that all variables that appear as exponents represent positive integers. $$x^{2 a}+2 x^{a}-24$$

Solve each equation. You will need to use the factoring techniques that we discussed throughout this chapter. $$6 x^{2}+25 x+14=0$$

Set up an equation and solve each problem. The perimeter of a rectangle is 32 inches, and the area is 60 square inches. Find the length and width of the rectangle.

Solve each equation. You will need to use the factoring techniques that we discussed throughout this chapter. $$28 n^{2}-47 n+15=0$$

Solve each equation. You will need to use the factoring techniques that we discussed throughout this chapter. $$s^{2}-4 s=21$$
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