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Chapter 16: Problem 36

For Exercises 7 to \(47,\) solve by completing the square. $$y^{2}+y-4=0$$

Short Answer

Expert verified
The solution to the equation \(y^{2} + y - 4 = 0\) is \(y = - 1/2 + \frac{\sqrt{17}}{2}\) or \(y = - 1/2 - \frac{\sqrt{17}}{2}\)

Step by step solution

01

Rearrange the term

Rearrange the equation as follows: \(y^{2} + y = 4\). Then complete the square, for a quadratic in the form \(y^{2} + b * y\), the term added to both sides would be \((b/2)^{2}\). In this case, \(b = 1\), so add \((1/2) ^{2}\) to both sides of the equation, which results in \(y^{2} + y + (1/2)^{2} = 4 + (1/2)^{2}\)
02

Simplify the square

Simplify the equation to obtain \( (y + 1/2)^{2} = 4 + 1/4\), which simplifies to \( (y + 1/2)^{2} = 17/4\)
03

Solve for y

Apply the square root property to the equation. That is, if \( x^{2} = c\), then \( x = \sqrt{c}\) or \(x = - \sqrt{c}\). Thus we obtain \(y + 1/2 = \sqrt{17/4}\) or \(y + 1/2 = - \sqrt{17/4}\)
04

Finalize the solution

Express the quantities \(\sqrt{17/4}\) as \(\pm \frac{\sqrt{17}}{2}\) ,then on solving each equation for \(y\), we get \(y = - 1/2 + \frac{\sqrt{17}}{2}\) or \(y = - 1/2 - \frac{\sqrt{17}}{2}\)

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Key Concepts

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

Understanding Quadratic Equations
Quadratic equations are fundamental in algebra and are characterized by an equation that can be written in the form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown variable. These equations form a parabola when graphed on a coordinate plane.
Understanding the significance of each term is key. The \( ax^2 \) is the quadratic term, \( bx \) is the linear term, and \( c \) is the constant term. Each has a specific role in determining the shape and position of the parabola.
  • The quadratic term dictates the direction (upward or downward) and width of the parabola.
  • The linear term affects the symmetry and horizontal shift of the graph.
  • The constant term shifts the parabola vertically.
Completing the square is a technique that transforms a quadratic equation into a perfect square format, making it easier to solve.
Techniques for Solving Equations by Completing the Square
Solving equations by completing the square is a valuable method when dealing with quadratic equations. This is particularly useful when factoring is complex or not possible. The main goal is to form a perfect square trinomial, allowing for easy extraction of solutions.
Here's a straightforward procedure to complete the square:
  • First, ensure the quadratic term, \( y^2 \), is isolated with the linear term on one side of the equation.
  • Next, adjust the equation as \( y^2 + by = c \).
  • Identify the coefficient of \( y \), or \( b \), and calculate \( (b/2)^2 \). Add this value to both sides to maintain equality.
  • Rewrite the equation with the completed square: \( (y + b/2)^2 = c + (b/2)^2 \).
This method provides a clean pathway to solve for \( y \) using further algebraic techniques.
Applying the Square Root Property
The square root property is a crucial step in solving quadratic equations transformed through completing the square method. Once the equation is in the form \((y + k)^2 = m\), you can efficiently find solutions by leveraging this property.
The square root property states that if \( x^2 = c \), then the solution is \( x = \sqrt{c} \) or \( x = -\sqrt{c} \).
  • Apply the square root to both sides of the equation: \( y + 1/2 = \pm \sqrt{17/4} \).
  • Solve for the variable by isolating \( y \). In this equation, subtracting \( 1/2 \) from both sides gives the solutions: \( y = -1/2 + \frac{\sqrt{17}}{2} \) and \( y = -1/2 - \frac{\sqrt{17}}{2} \).
This method not only simplifies solving but ensures you account for both possible solutions, which is critical given the nature of quadratic equations.

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Most popular questions from this chapter

In a slow-pitch softball game, the height of the ball thrown by a pitcher can be modeled by the equation \(h=-16 t^{2}+24 t+4,\) where \(h\) is the height of the ball in feet and \(t\) is the time, in seconds, since it was released by the pitcher. If the batter hits the ball when it is 2 ft off the ground, for how many seconds has the ball been in the air? Round to the nearest hundredth. (PICTURE NOT COPY)

For a quadratic equation of the form \(x^{2}+b x+c=0,\) the sum of the solutions is equal to the opposite of \(b\), and the product of the solutions is equal to \(c .\) For example, the solutions of the equation \(x^{2}+5 x+6=0\) are \(-2\) and \(-3 .\) The sum of the solutions is \(-5,\) the opposite of the coefficient of \(x\). The product of the solutions is \(6,\) the constant term. This is one way to check the solutions of a quadratic equation. Use this method to determine whether the given numbers are solutions of the equation. If they are not solutions of the equation, find the solutions. $$x^{2}-8 x-14=0 ;-4+\sqrt{15} \text { and }-4-\sqrt{15}$$

Solve by taking square roots. $$z^{2}+49=0$$

What is the \(y\) -intercept of the parabola with equation \(y=a x^{2}+b x + c ?\)

Graph. (GRAPH CANNOT COPY) $$f(x)=2 x^{2}-1$$
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