Problem 43
Question
Solve the equations using any method you choose. $$2 y^{2}-50=0$$
Step-by-Step Solution
Verified Answer
The solutions are \(y = 5\) and \(y = -5\).
1Step 1: Isolating the Quadratic Term
To solve the equation, our first step is to isolate the quadratic term. Start with the equation: \(2y^{2} - 50 = 0\). Add 50 to both sides to get: \(2y^{2} = 50\).
2Step 2: Simplifying the Coefficient
Now, divide both sides of the equation by 2 to simplify and get the coefficient of \(y^{2}\) to be 1. This gives us: \(y^{2} = 25\).
3Step 3: Solving for y by Taking the Square Root
To solve for \(y\), take the square root of both sides of the equation. Remember, taking the square root results in both a positive and a negative solution: \(y = \pm \sqrt{25}\). Therefore, \(y = 5\) or \(y = -5\).
Key Concepts
Isolating the Quadratic TermTaking the Square RootPositive and Negative Solutions
Isolating the Quadratic Term
When you come across a quadratic equation, your first task is often to isolate the quadratic term. This simply means getting one side of the equation to have only the term with the variable squared. In our example, we start with the equation: \(2y^{2} - 50 = 0\).
The term with the variable is \(2y^{2}\), and by moving the constant term 50 to the other side of the equation, you essentially isolate \(2y^{2}\). Add 50 to both sides so that the equation becomes \(2y^{2} = 50\). Now, the quadratic expression \(2y^{2}\) stands alone on the left side.
This step is crucial as it simplifies your equation and sets the stage for further operations. Isolating the expression helps in understanding what's affecting the variable and is vital for accurately solving the quadratic equation.
The term with the variable is \(2y^{2}\), and by moving the constant term 50 to the other side of the equation, you essentially isolate \(2y^{2}\). Add 50 to both sides so that the equation becomes \(2y^{2} = 50\). Now, the quadratic expression \(2y^{2}\) stands alone on the left side.
This step is crucial as it simplifies your equation and sets the stage for further operations. Isolating the expression helps in understanding what's affecting the variable and is vital for accurately solving the quadratic equation.
Taking the Square Root
Once the quadratic term is isolated and simplified to \(y^{2} = 25\), the next step is to take the square root of both sides. This allows us to solve for the variable \(y\) directly.
The act of taking the square root is straightforward: when you have \(y^{2}\), its square root will yield \(y\), assuming you've simplified correctly. For \(y^{2} = 25\), by taking the square root, you solve \(y = \pm \sqrt{25}\).
It's important to remember that taking a square root can often seem intimidating, but it becomes simpler when you recognize the perfect squares. For instance, \(\sqrt{25}\) is a basic calculation that results in 5. The use of square roots isn't just for computational purposes, but also to express the potential solutions as simply as possible.
The act of taking the square root is straightforward: when you have \(y^{2}\), its square root will yield \(y\), assuming you've simplified correctly. For \(y^{2} = 25\), by taking the square root, you solve \(y = \pm \sqrt{25}\).
It's important to remember that taking a square root can often seem intimidating, but it becomes simpler when you recognize the perfect squares. For instance, \(\sqrt{25}\) is a basic calculation that results in 5. The use of square roots isn't just for computational purposes, but also to express the potential solutions as simply as possible.
Positive and Negative Solutions
Quadratic equations are special because they usually have two solutions, one positive and one negative. This is especially true when the solutions are derived from taking a square root.
In the case of our problem: \(y = \pm \sqrt{25}\), the \(\pm\) symbol represents both positive and negative outcomes. Therefore, you have \(y = 5\) or \(y = -5\).
The nature of square roots produces these two solutions because when you square either \(5\) or \(-5\), you return to the original squared term 25. This is a beautiful demonstration of symmetry in mathematics, providing a complete picture of all possible solutions for an equation with a squared term. Recognizing that quadratic equations often yield two solutions is a fundamental part of understanding how these equations behave.
In the case of our problem: \(y = \pm \sqrt{25}\), the \(\pm\) symbol represents both positive and negative outcomes. Therefore, you have \(y = 5\) or \(y = -5\).
The nature of square roots produces these two solutions because when you square either \(5\) or \(-5\), you return to the original squared term 25. This is a beautiful demonstration of symmetry in mathematics, providing a complete picture of all possible solutions for an equation with a squared term. Recognizing that quadratic equations often yield two solutions is a fundamental part of understanding how these equations behave.
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Problem 43
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