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Chapter 6: Problem 15

Solve the equation for \(x\) and then solve it for \(y .\) $$ x-y^{2}=5 $$

Short Answer

Expert verified
For x: \(x = y^2 + 5\). For y: \(y = \pm \sqrt{x - 5}\).

Step by step solution

01

Solving for x

To solve for \(x\), we start with the given equation: \(x - y^2 = 5\). Our goal is to isolate \(x\) on one side of the equation. We can do this by adding \(y^2\) to both sides of the equation. This gives us: \(x = y^2 + 5\). Thus, the solution for \(x\) in terms of \(y\) is \(x = y^2 + 5\).
02

Solving for y

Next, to solve for \(y\), we will rearrange the equation into the standard form of a quadratic equation: \(y^2 = x - 5\). We solve this by taking the square root of both sides. Remember that taking the square root introduces a positive and negative solution. So, \(y = \pm \sqrt{x - 5}\). Therefore, the solutions for \(y\) in terms of \(x\) are \(y = \sqrt{x - 5}\) and \(y = -\sqrt{x - 5}\).

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Key Concepts

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

Solving for Variables
Solving for variables is a foundational skill in algebra that involves expressing one variable in terms of others. In our exercise, we begin with the equation \( x - y^2 = 5 \). The main goal here is to "solve for \( x \)," meaning to isolate \( x \) on one side of the equation. Here's how you can do it:
  • Start with your given equation: \( x - y^2 = 5 \).
  • We want \( x \) by itself, so we add \( y^2 \) to both sides: \( x = y^2 + 5 \).
After rearranging, you have expressed \( x \) in terms of \( y \), which is \( x = y^2 + 5 \).
This process is essential for troubleshooting and understanding relationships between variables. It helps in seeing how changes in \( y \) affect \( x \). Let鈥檚 move on to solving for \( y \) next.
Quadratic Equations
Quadratic equations are equations that can be expressed in the form \( ax^2 + bx + c = 0 \). For our task, when rearranging to solve for \( y \), we encounter a quadratic equation in terms of \( y \). It looks like \( y^2 = x - 5 \). There are a few things you should remember about quadratic equations:
  • The highest exponent of the variable (in this case \( y \)) is 2, making it a quadratic expression.
  • Quadratic equations can have two solutions because they describe a parabola when graphed.
Why do we care about the graph? Knowing the shape and nature of quadratic equations helps us anticipate the number of solutions and their potential values. This will bring us to the "Square Root Method," which is one of the key ways to solve quadratics.
Square Root Method
The square root method is a technique used to solve quadratic equations, particularly when the equation is already simplified to the form \( y^2 = k \), where \( k \) is some constant. In our situation, after solving for \( y \), the equation becomes \( y^2 = x - 5 \). Here's how to use the square root method:
  • To isolate \( y \), take the square root of both sides: \( y = \pm \sqrt{x - 5} \).
  • Remember, whenever you take the square root in an equation, include both the positive and negative roots. This is why we have \( \pm \sqrt{x - 5} \).
This is how we get two solutions for \( y \): one being positive and one negative. The square root method is powerful because it simplifies solving quadratics when they can be transformed into \( y^2 = k \). By understanding and practicing this method, you can tackle a wide variety of quadratic problems.

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Most popular questions from this chapter

Graph the solution set to the system of inequalities. $$ \begin{aligned} &x^{2}+y^{2} \leq 4\\\ &x^{2}+2 y \leq 2 \end{aligned} $$

Evaluate the matrix expression. $$\left[\begin{array}{rrr}5 & -1 & 6 \\\\-2 & 10 & 12 \\\5 & 2 & 9\end{array}\right]-\left[\begin{array}{rrr}-1 & 2 & 2 \\\2 & -1 & 2 \\\2 & 2 & -1\end{array}\right]$$

Graph the solution set to the system of inequalities. $$ \begin{aligned} &x^{2}+2 y \leq 4\\\ &x^{2}-y \leq 0 \end{aligned} $$

Digititing Letters Complete the following. (a) Design a matrix A with dimension \(4 \times 4\) that represents a digital image of the given letter. Asswme that there are four gray levels from 0 to 3 . (b) Find a matrix \(B\) such that \(B-A\) represents the negative image of the picture represented by matrix \(\mathbf{A}\) from part ( \(a\) ). $$\mathbf{z}$$

Properties of Matrices Use a graphing calculator to evaluate the expression with the given matrices \(A, B,\) and \(C .\) Compare your answers for parts (a) and (b). Then interpret the results. $$A=\left[\begin{array}{rrr}2 & -1 & 3 \\\1 & 3 & -5 \\\0 & -2 & 1\end{array}\right], B=\left[\begin{array}{rrr}6 & 2 & 7 \\\3 & -4 & -5 \\\7 & 1 & 0\end{array}\right]$$ $$C=\left[\begin{array}{lll}1 & 4 & -3 \\\8 & 1 & -1 \\\4 & 6 & -2\end{array}\right]$$ (a) \((A-B)^{2}\) (b) \(A^{2}-A B-B A+B^{2}\)
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