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Solve each quadratic equation by completing the square. Leave your answer in radical form. a. \(x^{2}-4 x-8=0\) (a) b. \(x^{2}+2 x-1=-5\) c. \(x^{2}+10 x-9=0\) d. \(5 x^{2}+10 x-7=28\)

Short Answer

Expert verified
(a) \(x = 2 \pm 2\sqrt{3}\); (b) \(x = 0, -2\); (c) \(x = -5 \pm \sqrt{34}\); (d) \(x = -1 \pm 2\sqrt{2}\).

Step by step solution

01

Rearrange the Equation

For equation [a] \( x^2 - 4x - 8 = 0 \), first add 8 to both sides to get \( x^2 - 4x = 8 \).
02

Complete the Square

Take half of the coefficient of \( x \), square it, and add it to both sides. The coefficient of \( x \) is \(-4\), half of it is \(-2\), and squaring it gives \(4\). Add \(4\) to both sides: \(x^2 - 4x + 4 = 8 + 4 \).
03

Form a Perfect Square Trinomial

Write the left-hand side as a square: \((x - 2)^2 = 12\).
04

Solve for x using the Square Root Method

Take the square root of both sides: \(x - 2 = \pm \sqrt{12}\). Simplify the square root: \(x - 2 = \pm 2\sqrt{3}\).
05

Solve for Final Values of x

Add 2 to both sides to solve for \(x\): \(x = 2 \pm 2\sqrt{3}\).
06

(Part b): Simplify the Equation

For equation [b] \(x^2 + 2x - 1 = -5\), simplify the equation to \(x^2 + 2x + 4 = 0\) by adding \(5\) on both sides.
07

(Part b): Complete the Square

Take half of the coefficient of \(x\) (which is \(1\)), square it (giving \(1\)), and add it inside the equation: \(x^2 + 2x + 1 = 1\).
08

(Part b): Form a Perfect Square Trinomial

The expression is now a perfect square trinomial \((x+1)^2 = 1\).
09

(Part b): Solve for x using the Square Root Method

Take the square root of both sides: \(x + 1 = \pm 1\).
10

(Part b): Solve for x

Subtract 1 from both sides: \(x = -1 \pm 1\). Thus, \(x = 0\) or \(x = -2\).
11

(Part c): Rearrange the Equation

For equation [c] \(x^2 + 10x - 9 = 0\), add \(9\) to both sides: \(x^2 + 10x = 9\).
12

(Part c): Complete the Square

Take half of the coefficient of \(x\) (which is \(5\)), square it (giving \(25\)), and add to both sides: \(x^2 + 10x + 25 = 9 + 25\).
13

(Part c): Form a Perfect Square Trinomial

The left side is a perfect square trinomial: \((x + 5)^2 = 34\).
14

(Part c): Solve for x using the Square Root Method

Take the square root of both sides: \(x + 5 = \pm \sqrt{34}\).
15

(Part c): Solve for x

Subtract \(5\) from both sides: \(x = -5 \pm \sqrt{34}\).
16

(Part d): Simplify the Equation

For equation [d] \(5x^2 + 10x - 7 = 28\), add \(7\) to \(28\) to simplify, moving \(-7\) to the right side: \(5x^2 + 10x = 35\).
17

(Part d): Factor out the leading Coefficient

Factor out \(5\) from the left side: \(5(x^2 + 2x) = 35\). Divide by \(5\): \(x^2 + 2x = 7\).
18

(Part d): Complete the Square

Take half of \(2\), square it, and add it to both sides: \(x^2 + 2x + 1 = 8\).
19

(Part d): Form a Perfect Square Trinomial

The expression \(x^2 + 2x + 1 = 8\) can be written as: \((x + 1)^2 = 8\).
20

(Part d): Solve for x using the Square Root Method

Take the square root of both sides of the equation: \(x + 1 = \pm \sqrt{8}\). This simplifies to \(x + 1 = \pm 2\sqrt{2}\).
21

(Part d): Solve for Final Values of x

Subtract 1 from both sides: \(x = -1 \pm 2\sqrt{2}\).

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Key Concepts

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

Quadratic Equations
Quadratic equations are a type of polynomial equation that has the form \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). These equations are important in algebra due to their wide range of applications, from physics to finance. The graph of a quadratic equation is a parabola, which can open upwards or downwards depending on the sign of \(a\). The solutions of a quadratic equation correspond to the points where the graph intersects the x-axis.

There are several methods to solve quadratic equations, including factoring, using the quadratic formula, and completing the square. Completing the square is particularly useful when equations do not easily factor. It systematically rearranges and alters the equation to create a "perfect square trinomial," making it easier to solve using the square root method. Each method provides different insights and tools to approach quadratic problems, depending on the situation and equation structure.
Perfect Square Trinomial
A perfect square trinomial is a special kind of trinomial that can be written as the square of a binomial. In other words, an expression like \(x^2 + 2bx + b^2\) is a perfect square trinomial because it can be rewritten as \((x + b)^2\).

When solving quadratic equations by completing the square, our goal is to transform part of the equation into a perfect square trinomial. This involves a few key steps:
  • Isolate the quadratic and linear terms: Move the constant to the other side of the equation.
  • Add the square: Take half the coefficient of the linear term, square it, and add it to both sides of the equation.
  • Rewrite as a binomial square: This new trinomial is then expressed as the square of a binomial.
Turning a quadratic equation into a perfect square trinomial simplifies the process of finding the solutions because it directly sets up the equation for the square root method.
Square Root Method
The square root method is a straightforward technique used to solve equations that are set as a perfect square trinomial. Once an equation is in the form \((x \, \pm \, m)^2 = n\), the square root method allows us to take the square root of both sides.

Here's how the process works:
  • Isolate the square: Make sure your equation is set up as a perfect square on the left and a constant on the right.
  • Take the square root: Apply the square root to both sides of the equation, remembering to consider both the positive and negative roots. This gives us \(x \pm m = \pm \sqrt{n}\).
  • Solve for the variable: Isolate the variable \(x\) to find the solution(s).
This method is particularly useful in the completing the square process because it quickly leads us to the solutions, especially when they are non-integral or involve radicals. By handling the equation in a simple setup of a perfect square, the square root method helps bridge the transition from algebraic manipulation to numeric solutions.
Radical Form Solutions
Solutions expressed in radical form include square roots and are not simplified into decimal form. These solutions are particularly common in cases where exact values are necessary, such as in theoretical mathematics or when working with irrational numbers.

When solving quadratic equations by completing the square, we often find that radical solutions arise when we apply the square root method. For instance, if the equation after completing the square is \((x - 2)^2 = 12\), solving it yields \(x - 2 = \pm \sqrt{12}\), which is simplified to \(x = 2 \pm 2\sqrt{3}\).

Expressing solutions in radical form is beneficial because:
  • It maintains the exactness of the solution.
  • It avoids the approximation errors that occur with decimal solutions.
  • It is essential when dealing with irrational numbers, where the decimal representation would be infinite and non-repeating.
In practice, keeping solutions in radical form ensures precision and accuracy, which can be critical in both academic settings and real-world applications.

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Most popular questions from this chapter

Find each product. Check your answers by using calculator tables or graphs. a. \((x+1)\left(2 x^{2}+3 x+1\right)\) b. \((2 x-5)\left(3 x^{2}+2 x-4\right)\)

This parabola has \(x\)-intercepts 3 and \(-2\). What is the equation of the line of symmetry? What is the \(x\)-coordinate of the vertex?

Sometimes you can spot the factors of a polynomial expression without a graph of the equation. The easiest factors to see are those called common monomial factors. If you can divide each term by the same expression, then there is a common factor. Factor each expression by removing the largest possible common monomial factor. a. \(4 x^{2}+12 x\) (a) b. \(6 x^{2}-4 x\) c. \(14 x^{4}+7 x^{2}-21 x\) (a) d. \(12 x^{5}+6 x^{3}+3 x^{2}\)

The Yo-yo Warehouse uses the equation \(y=-85 x^{2}+552.5 x\) to model the relationship between income and price for one of its top-selling yo-yos. In this model, \(y\) represents income in dollars and \(x\) represents the selling price in dollars of one item. a. Graph this relationship on your calculator, and describe a meaningful domain and range for this situation. (a) b. Describe a method for finding the vertex of the graph of this relationship. What is the vertex? c. What are the real-world meanings of the coordinates of the vertex? d. What is the real-world meaning of the two \(x\)-intercepts of the graph? e. Interpret the meaning of this model if \(x=5\).

In this exercise you will discover whether knowing the \(x\)-intercepts determines a unique quadratic equation. Work through the steps in 9 a-e to find an answer. Graph each equation to check your work. a. Write an equation of a parabola with \(x\)-intercepts at \(x=3\) and \(x=7\). b. Name the vertex of the parabola in 9 a. c. Modify your equation in 9 a so that the graph is reflected across the \(x\)-axis. Where are the \(x\)-intercepts? Where is the vertex? d. Modify your equation in 9 a to apply a vertical stretch with a factor of 2 . Where are the \(x\)-intercepts? Where is the vertex? (a) e. How many quadratic equations do you think there are with \(x\)-intercepts at \(x=3\) and \(x=7 ?\) How are they related to one another?

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