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Chapter 10: Problem 12

Solve the equation. Tell which method you used. \(y^{2}-4 y-5=0\)

Short Answer

Expert verified
The solutions of the equation \(y^{2}-4 y-5=0\) are \(y = 5\) and \(y = -1\). This solution uses the Quadratic Formula method.

Step by step solution

01

Identify the Parameters a, b, and c

In the equation \(y^{2}-4 y-5=0\), a = 1, b = -4, and c = -5
02

Input the Values into the Quadratic Formula

Plug the values of a, b, and c into the quadratic formula, which results in \(y = \frac{-(-4) \pm \sqrt{(-4)^{2}-4*1*(-5)}}{2*1}\)
03

Simplify the Expressions within the Square Root and the Numerator

Simplify the calculations within the square root and the numerator. This results in \(y = \frac{4 \pm \sqrt{16+20}}{2}\). Then, simplify under the square root, which yields \(y = \frac{4 \pm \sqrt{36}}{2}\)
04

Solve for y

Solve the equation to get the final results. The square root of 36 is 6. So, two solutions can be found: \(y = \frac{4+6}{2}\) or \(y = \frac{4-6}{2}\). These give us y = 5 and y = -1

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Key Concepts

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

Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations. A quadratic equation takes the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. The formula itself is given by:
  • \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
This formula is derived from the process of completing the square and can be used to find the roots of any quadratic equation.

One of the key features of the quadratic formula is the discriminant, \(b^2 - 4ac\). This part under the square root is crucial because it determines the nature of the roots (solutions):
  • If it is positive, there are two distinct real roots.
  • If it is zero, there is exactly one real root.
  • If it is negative, the roots are complex (not real).
Learning how to use the quadratic formula efficiently can simplify the process of finding solutions to quadratic equations.
Solving Equations
Solving equations involves finding the values of the variable that make the equation true. In the case of quadratic equations, you have several methods to solve them, such as:
  • Factoring: This method involves expressing the quadratic in a product form like \((x - p)(x - q) = 0\) if possible.
  • Completing the Square: This involves rearranging the equation into a perfect square trinomial on one side.
  • Quadratic Formula: As discussed earlier, it is a universal method that works when other methods become challenging.
Using the quadratic formula is particularly beneficial when the quadratic equation does not easily factor, such as in the example \(y^2 - 4y - 5 = 0\). Recognizing which method to apply depends on the particular equation and which approach will solve it most efficiently.
Roots of a Quadratic Equation
The roots of a quadratic equation are the values of the variable that satisfy the equation. For \(y^2 - 4y - 5 = 0\), the roots are the solutions found for \(y\) using the quadratic formula.
  • Start by identifying the coefficients \(a = 1\), \(b = -4\), and \(c = -5\).
  • Plug these into the formula: \(y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times (-5)}}{2 \times 1}\).
After simplifying, you calculate the expression under the square root, the discriminant, which is 36 in this case. The square root of 36 is 6, leading to:
  • \(y = \frac{4 + 6}{2} = 5\)
  • \(y = \frac{4 - 6}{2} = -1\)
These values, \(y = 5\) and \(y = -1\), are the roots of the quadratic equation. Understanding how to find the roots is critical as it represents solving the equation and finding what makes it true.

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Most popular questions from this chapter

Decide whether or not the ordered pair is a solution of the system of linear equations. $$\begin{aligned} &2 x+6 y=22\\\ &-x-4 y=-13 \quad(-5,-2) \end{aligned}$$

Solve the equation. \(|x+6|=13\)

Use factoring to solve the equation. Use a graphing calculator to check your solution if you wish. $$90 x^{2}-120 x+40=0$$

Find the greatest common factor of the numbers. 15 and 45

$$x^{2}-\frac{5}{3} x+\frac{25}{36}=0$$ $$112 x^{2}-252=0$$
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