Problem 14 Solve the equation by using the ... [FREE SOLUTION] (2024)

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Chapter 11: Problem 14

Solve the equation by using the quadratic formula where appropriate. $$u^{2}-3 u+3=0$$

Short Answer

Expert verified

The solutions are \( u = \frac{3 + i\sqrt{3}}{2} \) and \( u = \frac{3 - i\sqrt{3}}{2} \).

Step by step solution

01

Identify coefficients

To solve the quadratic equation using the quadratic formula, identify the coefficients. For the equation \(u^2 - 3u + 3 = 0\), the coefficients are: \(a = 1\), \(b = -3\), and \(c = 3\).

03

Substitute the coefficients into the formula

Substitute the coefficients \(a\), \(b\), and \(c\) into the quadratic formula: \[ u = \frac{-(-3) \,\pm\, \sqrt{(-3)^2 - 4(1)(3)}}{2(1)} \]

04

Simplify inside the square root

Calculate the expression inside the square root (the discriminant): \[ (-3)^2 - 4(1)(3) = 9 - 12 = -3 \] So the equation becomes: \[ u = \frac{3 \,\pm\, \sqrt{-3}}{2} \]

05

Write the solution in terms of imaginary numbers

Since the discriminant is negative, the solutions will be complex numbers. Recall that \( \sqrt{-3} = i\sqrt{3} \), where \( i \) is the imaginary unit. Therefore, the equation becomes: \[ u = \frac{3 \,\pm\, i\sqrt{3}}{2} \]

06

Separate the solutions

The solutions to the equation are: \[ u = \frac{3 + i\sqrt{3}}{2} \] and \[ u = \frac{3 - i\sqrt{3}}{2} \]

Key Concepts

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

solving quadratic equations

A quadratic equation is any equation that can be written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Solving these equations means finding values for the variable (typically represented as \(x\)) that satisfy the equation. One reliable method for solving quadratic equations is using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
To use this formula:

  • Identify the coefficients \(a\), \(b\), and \(c\) from the equation.
  • Substitute these values into the formula.
  • Simplify to find the solutions.

This formula works for any quadratic equation, whether the solutions are real or complex numbers.

complex numbers

Complex numbers are numbers that have both a real part and an imaginary part. They are written in the form \(a + bi\), where \(a\) is the real part, and \(bi\) is the imaginary part, with \(i\) representing the imaginary unit.
Complex numbers are especially useful in solving equations where the discriminant is negative, as they allow us to express solutions that aren't purely real. In our example equation \(u^2 - 3u + 3 = 0\), the solutions turned out to be complex numbers.
Complex numbers follow special rules for addition, subtraction, multiplication, and division:

  • Addition/Subtraction: Combine like terms (real with real, imaginary with imaginary).
  • Multiplication: Use the distributive property and remember that \(i^2 = -1\).
  • Division: Multiply the numerator and denominator by the complex conjugate to eliminate the imaginary part in the denominator.
discriminant in quadratic equations

The discriminant of a quadratic equation \(ax^2 + bx + c = 0\) is the expression \(b^2 - 4ac\) found under the square root in the quadratic formula.
It provides key information about the nature of the roots of the equation:

  • If the discriminant is positive \((b^2 - 4ac > 0)\), there are two distinct real roots.
  • If the discriminant is zero \((b^2 - 4ac = 0)\), there is one real root, or a repeated root.
  • If the discriminant is negative \((b^2 - 4ac < 0)\), there are two complex roots.

For our example, the discriminant was \(-3\), which indicated that the solutions would be complex numbers.
Understanding the discriminant can quickly tell you what type of solutions to expect before even solving the equation.

imaginary unit

The imaginary unit, denoted as \(i\), is defined by the property \(i^2 = -1\).
This concept allows us to expand the set of real numbers to include solutions to equations that don’t have real solutions. For instance, the equation \(x^2 + 1 = 0\) has no real solutions since there is no real number whose square is \(-1\).
However, using the imaginary unit, we can write the solutions as \(i\) and \(-i\). In our example, when we encountered \(\sqrt{-3}\), we expressed it as \(i\sqrt{3}\).
This standardized way of dealing with square roots of negative numbers simplifies the process of solving quadratic equations with negative discriminants.
The imaginary unit is foundational in complex numbers and significantly extends the possibilities for solutions in various mathematical contexts.

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Problem 14 Solve the equation by using the ... [FREE SOLUTION] (3)

Most popular questions from this chapter

Sketch the graphs of \(y=x^{2}, y=\frac{1}{2} x^{2},\) and \(y=\frac{1}{3} x^{2}\)on the same coordinate system. How would you describe the effect thecoefficients \(\frac{1}{2}\) and \(\frac{1}{3}\) have on the graph of \(y=x^{2} ?\)Sketch the graph of the given equation. Find the intercepts; approximate tothe nearest tenth where necessary. $$y=(x-1)^{2}$$Simplify as completely as possible. (Assume \(x \geq 0 .)\) $$\frac{12}{\sqrt{5}-\sqrt{3}}$$In Exercises \(1-64\), solve each of the given equations. If the equation isquadratic, use the factoring or square root method. If the equation has noreal solutions, say so. $$\frac{3}{x-2}+\frac{7}{x+2}=\frac{x+1}{x-2}$$In Exercises \(1-64\), solve each of the given equations. If the equation isquadratic, use the factoring or square root method. If the equation has noreal solutions, say so. $$\left(x+\frac{1}{3}\right)^{2}=\frac{3}{5}$$
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Problem 14 Solve the equation by using the ... [FREE SOLUTION] (2024)
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