/*! 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 43 Solve the equations using any me... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the equations using any method you choose. $$2 y^{2}-50=0$$

Short Answer

Expert verified
The solutions are \(y = 5\) and \(y = -5\).

Step by step solution

01

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: \(2y^{2} = 50\).
02

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: \(y^{2} = 25\).
03

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 \(y = -5\).

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Key Concepts

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

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.
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.
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.

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Most popular questions from this chapter

You \(\%\) need to recall the following definitions and results from elementary geometry. In a triangle, a line segment drawn from a vertex to the midpoint of the opposite side is called a median. The three medians of a triangle are concurrent; that is, they intersect in a single point. This point of intersection is called the centroid of the triangle. A line segment drawn from a vertex perpendicular to the opposite side is an altitude. The three altitudes of a triangle are concurrent; the point where the altitudes intersect is the orthocenter of the triangle. This exercise illustrates the fact that the altitudes of a triangle are concurrent. Again, we'll be using \(\triangle A B C\) with vertices \(A(-4,0), B(2,0),\) and \(C(0,6) .\) Note that one of the altitudes of this triangle is just the portion of the \(y\) -axis extending from \(y=0\) to \(y=6 ;\) thus, you won't need to graph this altitude; it will already be in the picture. (a) Using paper and pencil, find the equations for the three altitudes. (Actually, you are finding equations for the lines that coincide with the altitude segments.) (b) Use a graphing utility to draw \(\triangle A B C\) along with the three altitude lines that you determined in part (a). Note that the altitudes appear to intersect in a single point. Use the graphing utility to estimate the coordinates of this point. (c) Using simultaneous equations (from intermediate algebra), find the exact coordinates of the orthocenter. Are your estimates in part (b) close to these values?

(a) Graph the equation \(y=20 / x\) using a standard viewing rectangle. (b) Although both the \(x\) - and the \(y\) -axes are asymptotes for this curve, the graph in part (a) does not show this clearly. Take a second look, using a viewing rectangle that extends from -100 to 100 in both the \(x\) -and the \(y\) -directions. Note that the curve indeed appears indistinguishable from an asymptote when either \(|x|\) or \(|y|\) is sufficiently large.

Graph the equation after determining the \(x\) - and \(y\) -intercepts and whether the graph possesses any of the three types of symmetry described on page 58 $$y=|x|-2$$

Graph the equation after determining the \(x\) - and \(y\) -intercepts and whether the graph possesses any of the three types of symmetry described on page 58 $$x=y^{2}-1$$

You \(\%\) need to recall the following definitions and results from elementary geometry. In a triangle, a line segment drawn from a vertex to the midpoint of the opposite side is called a median. The three medians of a triangle are concurrent; that is, they intersect in a single point. This point of intersection is called the centroid of the triangle. A line segment drawn from a vertex perpendicular to the opposite side is an altitude. The three altitudes of a triangle are concurrent; the point where the altitudes intersect is the orthocenter of the triangle. This exercise provides an example of the fact that the medians of a triangle are concurrent. (a) The vertices of \(\triangle A B C\) are as follows: $$A(-4,0) \quad B(2,0) \quad C(0,6)$$ Use a graphing utility to draw \(\triangle A B C\). (since \(\overline{A B}\) coincides with the \(x\) -axis, you won't need to draw a line segment for this side.) Note: If the graphing utility you use does not have a provision for drawing line segments, you will need to determine an equation for the line in each case and then graph the line. (b) Find the coordinates of the midpoint of each side of the triangle, then include the three medians in your picture from part (a). Note that the three medians do appear to intersect in a single point. Use the graphing utility to estimate the coordinates of the centroid. (c) Using paper and pencil, find the equation of the medians from \(A\) to \(\overline{B C}\) and from \(B\) to \(\overline{A C}\). Then (using simultaneous equations from intermediate algebra), determine the exact coordinates of the centroid. How do these numbers compare with your estimates in part (b)?

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