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Chapter 3: Problem 37

Solve each equation. For equations with real solutions, support your answers graphically. $$x^{2}-2 x-4=0$$

Short Answer

Expert verified
The solutions are \( x = 1 + \sqrt{5} \) and \( x = 1 - \sqrt{5} \).

Step by step solution

01

Identify the Type of Equation

The given equation is a quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = -2 \), and \( c = -4 \).
02

Calculate the Discriminant

Use the formula for the discriminant \( D = b^2 - 4ac \). Substitute the values: \( D = (-2)^2 - 4 \times 1 \times (-4) = 4 + 16 = 20 \). The discriminant is positive, so there are two real solutions.
03

Solve the Quadratic Equation Using the Quadratic Formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{D}}{2a} \). Substitute the values: \( x = \frac{-(-2) \pm \sqrt{20}}{2 \times 1} = \frac{2 \pm \sqrt{20}}{2} \).
04

Simplify the Solutions

Simplify \( \sqrt{20} \) to \( 2\sqrt{5} \). The solutions are \( x = \frac{2 + 2\sqrt{5}}{2} \) and \( x = \frac{2 - 2\sqrt{5}}{2} \). Simplify further: \( x = 1 + \sqrt{5} \) and \( x = 1 - \sqrt{5} \).
05

Graphical Verification

Plot the quadratic equation \( y = x^2 - 2x - 4 \) on a graph. The curve will intersect the x-axis at the points \( x = 1 + \sqrt{5} \) and \( x = 1 - \sqrt{5} \). These are the x-intercepts and verify the real solutions graphically.

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Key Concepts

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

Discriminant
The discriminant plays a crucial role in determining the nature of the solutions for a quadratic equation. It is represented by the letter \( D \) and calculated using the formula \( D = b^2 - 4ac \). The values for \( b \), \( a \), and \( c \) are the coefficients from the equation in the standard form \( ax^2 + bx + c = 0 \).
Using the discriminant:
  • If \( D > 0 \), the quadratic equation has two distinct real solutions.
  • If \( D = 0 \), the equation has exactly one real solution, often referred to as a double root.
  • If \( D < 0 \), the equation has no real solutions; instead, it has two complex conjugate solutions.
By calculating the discriminant for our example equation \( x^2 - 2x - 4 = 0 \), we found that \( D = 20 \). This positive discriminant implies there are indeed two distinct real solutions for the equation.
Quadratic Formula
The quadratic formula is a cornerstone tool for finding the roots of any quadratic equation of the form \( ax^2 + bx + c = 0 \). The formula is given by\[x = \frac{-b \pm \sqrt{D}}{2a}\]where \( D \) is the discriminant \( b^2 - 4ac \).
To use this formula effectively:
  • Calculate the discriminant \( D \).
  • Substitute the values of \( a \), \( b \), and \( D \) into the quadratic formula.
  • Solve for \( x \) by computing the values derived from the formula, accounting for both \( + \) and \( - \).
Using the equation \( x^2 - 2x - 4 = 0 \), we applied the quadratic formula and obtained the solutions \( x = 1 + \sqrt{5} \) and \( x = 1 - \sqrt{5} \). These represent the two distinct real roots of the equation.
Graphical Solutions
Graphing a quadratic equation provides a visual representation of the solutions. The graph of a quadratic equation \( y = ax^2 + bx + c \) forms a parabola on the coordinate plane.
Key characteristics of the graph:
  • The x-intercepts of the parabola, where it crosses the x-axis, correspond to the roots of the equation.
  • A positive \( a \) value indicates the parabola opens upwards, while a negative \( a \) means it opens downwards.
  • The vertex of the parabola is the highest or lowest point, depending on the direction the parabola opens.
For the equation \( y = x^2 - 2x - 4 \), graphing confirms the real solutions are \( x = 1 + \sqrt{5} \) and \( x = 1 - \sqrt{5} \). At these x-values, the graph intersects the x-axis, providing a visual verification of our algebraic solutions.
Real Solutions
Real solutions refer to the values of \( x \) that satisfy the quadratic equation and can be plotted on a number line. These solutions are derived using the quadratic formula or by graphing the corresponding parabola.
Things to note about real solutions:
  • They are the x-intercepts or zeros of the quadratic function.
  • Real solutions can be rational or irrational numbers, depending on the value of the discriminant.
  • If a quadratic has real solutions, these can be verified graphically by observing where the parabola intersects the x-axis.
In our example of \( x^2 - 2x - 4 = 0 \), the real solutions \( x = 1 + \sqrt{5} \) and \( x = 1 - \sqrt{5} \) are indeed real and can be plotted, as evidenced by their presence at the points where the graph of the equation meets the x-axis.

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Most popular questions from this chapter

Draw by hand a rough sketch of the graph of each function. (You may wish to support your answer with a calculator graph.) $$\begin{aligned}P(x) &=2 x^{5}-10 x^{4}+x^{3}-5 x^{2}-x+5 \\\&=(x-5)\left(x^{2}+1\right)\left(2 x^{2}-1\right)\end{aligned}$$

Use the given zero to completely factor \(P(x)\) into linear factors. $$\text { Zero: } 1+i ; \quad P(x)=x^{4}-2 x^{3}+3 x^{2}-2 x+2$$

Divide. $$\frac{8 x^{3}+10 x^{2}-12 x-15}{2 x^{2}-3}$$

Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=x^{3}+9 x^{2}-7 x-63 ; \quad k=-9$$

Solve each formula for the indicated variable. Leave \(\pm\) in answers when appropriate. Assume that no denominators are \(0 .\) $$P=\frac{E^{2} R}{(r+R)^{2}} \text { for } R$$
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