/*! 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 6 Use the quadratic formula to sol... [FREE SOLUTION] | 91Ó°ÊÓ

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Use the quadratic formula to solve each equation. These equations have real number solutions only. See Examples I through 3. $$ y^{2}+10 y+25=0 $$

Short Answer

Expert verified
The solution is \( y = -5 \).

Step by step solution

01

Identify the quadratic equation

The given equation is in the form of a quadratic equation: \( y^2 + 10y + 25 = 0 \). This matches the standard form \( ax^2 + bx + c = 0 \) where \( a = 1 \), \( b = 10 \), and \( c = 25 \).
02

Recall the quadratic formula

The quadratic formula is \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \). We will use this formula to find the values of \( y \).
03

Substitute into the quadratic formula

Substitute \( a = 1 \), \( b = 10 \), and \( c = 25 \) into the quadratic formula: \( y = \frac{{-10 \pm \sqrt{{10^2 - 4 \cdot 1 \cdot 25}}}}{2 \cdot 1} \).
04

Simplify and solve

First, calculate the discriminant: \( b^2 - 4ac = 10^2 - 4 \times 1 \times 25 = 100 - 100 = 0 \). Since the discriminant is 0, there is exactly one real solution. Now compute \( y = \frac{{-10 \pm \sqrt{0}}}{2} = \frac{{-10}}{2} = -5 \).
05

Interpret the solution

The solution to the equation \( y^2 + 10y + 25 = 0 \) is \( y = -5 \). This means the graph of the equation touches the x-axis at the point \( y = -5 \).

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Key Concepts

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

Quadratic Equation
A quadratic equation is any equation that can be written in the form \( ax^2 + bx + c = 0 \). This is known as the standard form of a quadratic equation. The terms \( a \), \( b \), and \( c \) are constants, with \( x \) representing the variable. It is important that \( a \) is not zero because if \( a \) were zero, the equation would not be quadratic. Instead, it would be linear. Quadratic equations are polynomial equations of degree 2, characterized by the highest exponent, which in this case is 2. In our original exercise, the equation \( y^2 + 10y + 25 = 0 \) is already in the standard form, with \( a = 1 \), \( b = 10 \), and \( c = 25 \).
Understanding the structure of a quadratic equation helps in identifying different methods for solving, including factoring, completing the square, or using the quadratic formula. Here, we use the quadratic formula for an exact solution.
Quadratic Solution
Finding the solution to a quadratic equation can be accomplished using several methods, but the quadratic formula is one of the most powerful tools. The quadratic formula is \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \).
This single formula gives you the solutions for any quadratic equation. The \( \pm \) sign means that you might get two solutions, one by adding \( \sqrt{{b^2 - 4ac}} \) and the other by subtracting it. In the context of our problem, the variable \( y \) replaces \( x \), and we substitute the values \( a = 1 \), \( b = 10 \), and \( c = 25 \). Thus, the solution process involves substituting these values into the quadratic formula to find \( y \). Because our discriminant is zero (as you'll see below), the quadratic equation \( y^2 + 10y + 25 = 0 \) has one solution, which is \( y = -5 \). By understanding and applying the quadratic formula, you can efficiently solve most quadratic equations you encounter.
Discriminant
The discriminant is a special part of the quadratic formula and plays a crucial role in determining the nature of the solutions. It is expressed as \( b^2 - 4ac \). The discriminant reveals whether the solutions to the quadratic equation are real or complex, and whether they are distinct or repeated.
Consider these possible scenarios:
  • If \( b^2 - 4ac > 0 \), the equation has two distinct real solutions.
  • If \( b^2 - 4ac = 0 \), there is exactly one real solution, also known as a repeated or double root.
  • If \( b^2 - 4ac < 0 \), the solutions are complex and not real.
In our exercise, the discriminant calculation gives \( 10^2 - 4 \times 1 \times 25 = 100 - 100 = 0 \). Thus, the discriminant being zero indicates that there is exactly one real solution. This unique solution arises because the graph of the quadratic touches the x-axis at a single point.
Real Solutions
Real solutions refer to solutions of the quadratic equation that are real numbers. They are not complex or imaginary. A real solution means the equation's graph intersects the x-axis at least once. Depending on the discriminant:
  • If it is greater than zero, the graph intersects the x-axis at two different points, providing two distinct real solutions.
  • If it is zero, the graph just touches the x-axis at one point, providing one real solution. This was the case in our problem where \( y = -5 \) is the sole solution.
  • If the discriminant is less than zero, the graph does not intersect the x-axis at all, and thus, there are no real solutions.
Understanding the nature of real solutions helps in visualizing the graph of the quadratic equation and comprehending its interaction with the x-axis. In practical situations, real solutions could represent calculable values that can be used in further mathematical or physical interpretations.

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