Chapter 11: Problem 13
Solve each of the following quadratic equations, and check your solutions. $$t^{2}+6 t=-12$$
Short Answer
Expert verified
The solutions are \( t = -3 + i\sqrt{3} \) and \( t = -3 - i\sqrt{3} \).
Step by step solution
01
Move All Terms to One Side
First, we need to write the quadratic equation in the standard form, which is \[ ax^2 + bx + c = 0 \] For the given equation, \[ t^2 + 6t = -12 \]We add 12 to both sides to bring all terms to one side, getting:\[ t^2 + 6t + 12 = 0 \]
02
Identify a, b, and c
To apply the quadratic formula or complete the square, identify the coefficients:\[ a = 1, \, b = 6, \, c = 12 \]
03
Use the Quadratic Formula
The quadratic formula is:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute the values of \( a \), \( b \), and \( c \):\[ t = \frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times 12}}{2 \times 1} \]
04
Calculate the Discriminant
The discriminant \( b^2 - 4ac \) determines the nature of the roots:\[ 6^2 - 4 \times 1 \times 12 = 36 - 48 = -12 \]Since the discriminant is negative, the equation has two complex solutions.
05
Solve for t Using Complex Numbers
Continue solving using complex numbers:\[ t = \frac{-6 \pm \sqrt{-12}}{2} \]Let \( \sqrt{-12} = \sqrt{-1 \times 12} = \sqrt{-1} \times \sqrt{12} = i \sqrt{12} \)\[ t = \frac{-6 \pm i \sqrt{12}}{2} \]
06
Simplify the Expression
Further simplify \( \sqrt{12} \):\( \sqrt{12} = \sqrt{4 \times 3} = 2 \sqrt{3} \)Thus, the roots are:\[ t = \frac{-6 \pm i \cdot 2 \sqrt{3}}{2} \]\[ t = -3 \pm i \sqrt{3} \]
07
Verify the Solutions
Verify the roots by substituting them back into the original equation. Since these are complex roots arising from a negative discriminant, they must satisfy the equation in terms of equal magnitude and conjugate imaginary parts, which cannot be directly checked without complex arithmetic verification.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations. It provides an efficient way to find the solutions to any equation of the form \(ax^2 + bx + c = 0\). The formula itself is:
This method is especially useful when factoring is difficult or impossible, or when the equation itself does not seem to fit a recognizable pattern.
- \(\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
This method is especially useful when factoring is difficult or impossible, or when the equation itself does not seem to fit a recognizable pattern.
Complex Solutions
When solving quadratic equations, the term under the square root in the quadratic formula, known as the discriminant, can be negative. A negative discriminant indicates that the equation has complex solutions rather than real ones. In mathematical terms, these are solutions that involve the imaginary unit \(i\), where \(i = \sqrt{-1}\).
For example, when calculating \(\sqrt{-12}\), you can rewrite it as \(i\sqrt{12}\) because \(\sqrt{-1} = i\). Then further simplify \(\sqrt{12}\) to \(2\sqrt{3}\).
Consequently, having a negative discriminant, the solutions are not just numerical values but involve complex numbers, expressed as \(a + bi\) and \(a - bi\), showing both a real part \(a\) and an imaginary component \(bi\). This happens because the discriminant affects the type of solution you get.
For example, when calculating \(\sqrt{-12}\), you can rewrite it as \(i\sqrt{12}\) because \(\sqrt{-1} = i\). Then further simplify \(\sqrt{12}\) to \(2\sqrt{3}\).
Consequently, having a negative discriminant, the solutions are not just numerical values but involve complex numbers, expressed as \(a + bi\) and \(a - bi\), showing both a real part \(a\) and an imaginary component \(bi\). This happens because the discriminant affects the type of solution you get.
Discriminant
The discriminant is a pivotal value in the quadratic formula that greatly influences the type of solutions you'll find. It's calculated as \(b^2 - 4ac\) and can inform us of several conditions based on its value:
- If the discriminant is positive, expect two distinct real solutions.
- If the discriminant is zero, expect exactly one real solution (a repeated root).
- If the discriminant is negative, the solutions will be complex or imaginary.