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Chapter 11: Problem 37

Solve by completing the square. $$3 k^{2}-6 k+12=0$$

Short Answer

Expert verified
The solutions for the given quadratic equation are \(k = 1 + \sqrt{-3}\) and \(k = 1 - \sqrt{-3}\), which are complex solutions, as there are no real solutions in this case.

Step by step solution

01

Write the equation in standard form

The given equation is already in the standard quadratic form: \(3k^2 - 6k + 12 = 0\)
02

Get the coefficient of the leading term to be 1

Divide the entire equation by the coefficient of the leading term (3): \(\frac{3k^2 - 6k + 12}{3} = \frac{0}{3}\) \(k^2 - 2k + 4 = 0\)
03

Add and subtract the value that completes the square

Take half of the coefficient of the linear term (k), which is -2, and square it: \((-2/2)^2 = (-1)^2 = 1\). Add and subtract this number inside the parentheses to complete the square: \(k^2 - 2k + 1 - 1 + 4 = 0 \) \((k - 1)^2 - 1 + 4 = 0\)
04

Solve for the variable

Simplify the equation and solve for k: \((k - 1)^2 = 1 - 4\) \((k - 1)^2 = -3\) Since the square of a real number cannot be negative, there are no real solutions to this equation. However, there can be complex solutions: \(k - 1 = \pm \sqrt{-3}\) \(k = 1 \pm \sqrt{-3}\) The solutions for the given equation are: \(k = 1 + \sqrt{-3}\) and \(k = 1 - \sqrt{-3}\)

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Key Concepts

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

Completing the Square
When solving a quadratic equation, one effective method is completing the square. It is especially useful when the equation doesn't factorize neatly. The goal here is to transform part of your quadratic into a perfect square trinomial, which simplifies the expression.
Completing the square involves a few straightforward steps:
  • First, ensure the coefficient of the square term is 1. If not, divide the entire equation by that coefficient.
  • Next, focus on the linear term. Take half of its coefficient and square the result. This squared number is the key to completing the square.
  • Add and subtract this number within the equation, creating a perfect square trinomial.
By doing this, you rewrite the quadratic in a form that鈥檚 easier to solve, especially when isolating the variables for solutions.
Complex Solutions
Quadratic equations traditionally have two solutions. Sometimes, these solutions are complex rather than real. A complex solution is needed when you encounter a negative number inside a square root, as the square of a real number cannot be negative.
Here's how to deal with this situation:
  • Identify the term within the square that led to a negative. This indicates complex solutions.
  • The imaginary unit, \(i\), is defined as \(i = \sqrt{-1}\). Use this to express the square root of a negative number.
  • In the case of \(\sqrt{-3}\), rewrite it using \(i\): \(\sqrt{-3} = i\sqrt{3}\).
Thus, complex solutions for quadratic equations occur when dealing with negative square roots, resulting in expressions involving imaginary numbers.
Standard Form
Quadratic equations are often presented in standard form, which is \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) represent constants, and the term with \(x^2\) is the quadratic or leading term.
Using the standard form offers several advantages:
  • It allows for uniform methods of solving the equation, such as factoring, completing the square, or using the quadratic formula.
  • Being in standard form makes it easier to identify the coefficients, which are critical for solving using various methods.
  • It provides a clear structure for manipulating and rearranging terms to facilitate problem-solving.
Overall, starting with a quadratic in its standard form can simplify the process of finding solutions and understanding the nature of the roots.

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Most popular questions from this chapter

Identify the conjugate of each complex number, then multiply the number and its conjugate. $$-6+4 i$$

Solve. An object is thrown upward from a height of \(24 \mathrm{ft}\). The height \(h\) of the object (in feet) \(t\) sec after the object is released is given by \(h=-16 t^{2}+24 t+24\) a) How long does it take the object to reach a height of \(8 \mathrm{ft} ?\) b) How long does it take the object to hit the ground?

Divide. Write the result in the form \(a+b i\). $\frac{2 i}{-3+7 i}$$

Solve. \(d^{-2}+d^{-1}-6=0\)

SOLVE. $$ 2=14+(g+4)^{2} $$
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