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Chapter 6: Problem 41

Solve each equation. $$ 9 y^{2}-49=0 $$

Short Answer

Expert verified
The solutions are \(y = \frac{7}{3}\) and \(y = -\frac{7}{3}\).

Step by step solution

01

Identify the Equation Type

The given equation is \(9y^{2}-49=0\). It is a quadratic equation in the standard form of \(ax^2+bx+c=0\), where \(a=9\), \(b=0\), and \(c=-49\).
02

Move the Constant to the Other Side

Rewrite the equation by moving the constant term to the other side: \(9y^2 = 49\).
03

Divide by the Coefficient of \(y^2\)

Divide both sides by 9 to isolate \(y^2\): \(y^2 = \frac{49}{9}\).
04

Apply the Square Root

Take the square root of both sides to solve for \(y\): \(y = \pm \sqrt{\frac{49}{9}}\).
05

Simplify the Square Root

Simplify \(\sqrt{\frac{49}{9}}\) by finding the square root of the numerator and the denominator separately: \(y = \pm \frac{\sqrt{49}}{\sqrt{9}}\). Since \(\sqrt{49} = 7\) and \(\sqrt{9} = 3\), we have \(y = \pm \frac{7}{3}\).

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Key Concepts

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

Standard Form of Quadratic Equation
The standard form of a quadratic equation is expressed as \(ax^2 + bx + c = 0\). This is a fundamental structure used when dealing with quadratic equations. Each component of the equation has a specific role:
  • \(a\) is the coefficient of \(x^2\), which defines the parabola's curvature.
  • \(b\) is the linear coefficient, influencing the location of the vertex horizontally.
  • \(c\) is the constant term, affecting the vertical position of the parabola.
To solve a quadratic equation, identifying these parameters is crucial as it helps in selecting the appropriate solution method. In the presented problem, \(9y^2 - 49 = 0\), the equation is in standard form with \(a = 9\), \(b = 0\), and \(c = -49\). Understanding the standard form allows us to manipulate the equation to find its roots using various methods.
Solving Quadratic Equations
Once a quadratic equation is identified, the next step is to solve it, which involves finding its roots or solutions. There are several methods to solve quadratic equations:
  • Factoring involves expressing the quadratic equation as a product of two binomials. However, not all equations are easily factorable.
  • The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), is a reliable method for any quadratic equation.
  • Completing the square makes the equation a perfect square trinomial.
In the given exercise, the equation \(9y^2 - 49 = 0\) is solved more straightforwardly by simplifying and using algebraic manipulation instead of factoring. After rearranging to \(9y^2 = 49\), isolating \(y^2\) helps set up for applying the square root method.
Square Root Method
The square root method is a highly efficient way to solve quadratic equations, especially when the equation is in the form \(ax^2 = c\). It's suitable when there's no linear term (\(b=0\)), as seen in our problem. Here's how it works:
  • First, isolate the squared term. In this equation, it begins as \(9y^2 = 49\).
  • Divide both sides by the coefficient of \(y^2\) (which is 9 in this case) to obtain \(y^2 = \frac{49}{9}\).
  • Apply the square root to both sides to solve for \(y\). This results in: \(y = \pm \sqrt{\frac{49}{9}}\).
  • Simplify the square root by separately finding the square roots of the numerator and the denominator: \(y = \pm \frac{7}{3}\).
The final solutions are \(y = \frac{7}{3}\) and \(y = -\frac{7}{3}\). This method is efficient and straightforward when perfectly matched to the equation at hand.

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Most popular questions from this chapter

The following is a list of random factoring problems. Factor each expression. If an expression is not factorable, write "prime." See Examples 1-5. $$ 8 x^{4}-8 $$

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Factor using rational numbers. $$ 24-x^{3}+8 x^{2}-3 x $$

The following is a list of random factoring problems. Factor each expression. If an expression is not factorable, write "prime." See Examples 1-5. $$ 25 x^{2}-16 y^{2} $$
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