Example Question - imaginary unit
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Solving a Quadratic Equation with Complex Solutions
La imagen muestra una ecuación que parece ser \( y + \frac{1}{y} = \frac{4}{3} \). Para resolver esta ecuación por \( y \), podemos seguir los siguientes pasos: 1. Multiplicar ambos lados de la ecuación por \( y \) para eliminar el denominador: \( y(y) + \frac{1}{y}(y) = \frac{4}{3}(y) \) 2. Esto simplifica a: \( y^2 + 1 = \frac{4}{3}y \) 3. Para resolver una ecuación cuadrática, queremos llevar todos los términos a un lado de la igualdad y establecer la ecuación igual a cero: \( y^2 - \frac{4}{3}y + 1 = 0 \) 4. Multiplicamos todo por 3 para eliminar los denominadores y obtener una ecuación con números enteros: \( 3y^2 - 4y + 3 = 0 \) 5. Ahora tenemos una ecuación cuadrática estándar y podemos usar la fórmula cuadrática o intentar factorizar. La ecuación parece no factorizarse fácilmente, entonces usaremos la fórmula cuadrática \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) donde \( a = 3 \), \( b = -4 \), y \( c = 3 \). 6. Sustituimos los valores en la fórmula cuadrática: \( y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(3)}}{2(3)} \) 7. Esto simplifica a: \( y = \frac{4 \pm \sqrt{16 - 36}}{6} \) \( y = \frac{4 \pm \sqrt{-20}}{6} \) 8. Como tenemos un número negativo dentro del radical, las soluciones serán números complejos. Extraemos el \(i\) (la unidad imaginaria) para representar \( \sqrt{-1} \): \( y = \frac{4 \pm \sqrt{20}i}{6} \) 9. Simplificamos \( \sqrt{20} \) como \( 2\sqrt{5} \) para obtener las soluciones en su forma más simplificada: \( y = \frac{4 \pm 2\sqrt{5}i}{6} \) 10. Finalmente, dividimos tanto el numerador como el denominador por 2: \( y = \frac{2 \pm \sqrt{5}i}{3} \) Entonces las soluciones son \( y = \frac{2}{3} + \frac{\sqrt{5}}{3}i \) y \( y = \frac{2}{3} - \frac{\sqrt{5}}{3}i \).
Interpretation of a Quadratic Equation with Multiple x² Values
La imagen muestra una ecuación matemática, la cual es: \[ y^{2} = x^{2} *\{-1, 3\} \] Esto indica que tenemos un conjunto con dos elementos al lado derecho de la ecuación: -1 y 3. Para resolver esta ecuación, hay que tomar en cuenta que vamos a trabajar con dos posibles valores para x², -1 y 3, y calcular los posibles valores de y² para cada uno de ellos. Primero, tomaremos x² = -1: \[ y^{2} = (-1) \] Sin embargo, aquí nos encontramos con un problema conceptual y matemático, ya que en los números reales, el cuadrado de un número jamás será negativo, porque tanto el cuadrado de un número positivo como de un número negativo siempre resulta en un número no negativo. Es decir, y² siempre será mayor o igual a cero en el conjunto de los números reales, por lo que no hay un número real cuyo cuadrado sea -1. Si estuviésemos trabajando con números complejos, podría decirse que \( y = i \) o \( y = -i \) donde \( i \) es la unidad imaginaria, que satisface \( i² = -1 \), pero en este caso no vamos a considerar soluciones complejas. Ahora tomaremos x² = 3: \[ y^{2} = 3 \] Aquí sí es posible encontrar valores reales para y, ya que buscamos dos números cuyos cuadrados sean 3. Estos números son \( \sqrt{3} \) y \( -\sqrt{3} \). Entonces, resumiendo, tenemos que: - Para x² = -1, no hay solución en los números reales. - Para x² = 3, las soluciones para y son \( y = \sqrt{3} \) y \( y = -\sqrt{3} \). Estos son los valores posibles de y con base en la ecuación presentada y el conjunto de valores para x² dado.
Understanding Complex Numbers and Inequalities
The image shows an inequality with a square root of a negative number: the square root of -86. In the real number system, the square root of a negative number is not defined because no real number squared gives a negative result. However, in the complex number system, the square root of a negative number involves the imaginary unit \( i \), where \( i^2 = -1 \). To express the square root of -86, we can factor out the imaginary unit \( i \), resulting in: \[ \sqrt{-86} = \sqrt{-1 \cdot 86} = \sqrt{-1} \cdot \sqrt{86} = i \sqrt{86} \] Since you're asked to place the square root of -86 within inequalities, it's important to note that complex numbers do not have a natural ordering like real numbers, so you cannot say that one complex number is greater than or less than another. Thus, the image prompts an operation which is not valid within the real number system and cannot be completed as a typical inequality. If we were to attempt to place this value in an inequality with real numbers, we could not do so meaningfully, as the complex number cannot be directly compared to real numbers in terms of being greater or lesser. However, the absolute value or magnitude of a complex number can be compared to real numbers. The magnitude of \( i\sqrt{86} \) is \( \sqrt{86} \), and we know that real numbers less than \( \sqrt{86} \) would be to the left of it on the real number line, and numbers greater than \( \sqrt{86} \) would be to the right if we were considering magnitude alone, ignoring the imaginary component. But without more context, and strictly speaking, the comparison symbols (<) in the image are not meaningful when applied to a complex number.
Understanding Complex Numbers and Inequalities
The expression in the image includes the square root of a negative number: \(-\sqrt{86}\). In the realm of real numbers, you cannot take the square root of a negative number, as square roots are defined only for non-negative numbers. However, in the context of complex numbers, the square root of a negative number is a multiple of the imaginary unit \(i\), where \(i\) is defined as \(i^2 = -1\). Thus, \(-\sqrt{86}\) in terms of complex numbers is \(-i\sqrt{86}\). This number is purely imaginary, and there is no real number less than or greater than it because comparisons of greater or less than do not apply to imaginary numbers. Therefore, the blank boxes on both sides of \(-\sqrt{86}\) cannot be filled with real numbers to create a true statement about inequalities. Filling these blanks would require a context where complex numbers are included, and even then, the notion of "greater than" or "less than" isn't meaningful in the same way as it is with real numbers.
Complex Number Multiplication
The image shows a handwritten problem which asks to compute the product of the following complex numbers: 1) \( Z_1 = 1 + 3i \) and \( Z_2 = 5 + i \) 2) \( Z_1 = 3 - 4i \) and \( Z_2 = 3 + 2i \) To compute the product of two complex numbers, you multiply them using the distributive property (also known as the FOIL method for binomials), which stands for "First, Outer, Inner, Last." Also remember that \( i^2 = -1 \). Let's solve the first product: \( Z_1 \cdot Z_2 = (1 + 3i)(5 + i) \) 1. First (1 * 5): 5 2. Outer (1 * i): i 3. Inner (3i * 5): 15i 4. Last (3i * i): 3i^2 Combining these we get: \( 5 + i + 15i + 3i^2 \) Since \( i^2 = -1 \), substitute for \( i^2 \): \( 5 + i + 15i + 3(-1) \) \( 5 + 16i - 3 \) Combine like terms: \( (5 - 3) + 16i \) \( 2 + 16i \) So the product of the first pair is \( 2 + 16i \). Now for the second product: \( Z_1 \cdot Z_2 = (3 - 4i)(3 + 2i) \) 1. First (3 * 3): 9 2. Outer (3 * 2i): 6i 3. Inner (-4i * 3): -12i 4. Last (-4i * 2i): -8i^2 Now combine these: \( 9 + 6i - 12i - 8i^2 \) Substitute \( i^2 \) with -1: \( 9 + 6i - 12i - 8(-1) \) \( 9 - 6i - 8 \) Combine like terms: \( (9 + 8) - 6i \) \( 17 - 6i \) So the product of the second pair is \( 17 - 6i \).
Multiplication of Complex Numbers
The image shows the multiplication of two complex numbers: \( (2 - i) \) and \( (1 + i) \), where \( i \) is the imaginary unit with the property that \( i^2 = -1 \). To solve this, we apply the distributive property (FOIL method) as we would with binomials: \[ (2 - i)(1 + i) = 2(1) + 2(i) - i(1) - i(i) \] Now we simplify each term: \[ = 2 + 2i - i - i^2 \] Remember that \( i^2 = -1 \), so the expression becomes: \[ = 2 + 2i - i + 1 \] Combine like terms: \[ = (2 + 1) + (2i - i) \] \[ = 3 + i \] So, the product of \( (2 - i) \) and \( (1 + i) \) is \( 3 + i \).
Solving Quadratic Equations with Complex Solutions
The equation given in the image is a quadratic equation of the form: \[ 2x^2 - 4x + 4 = 0 \] We can solve this equation by either factoring, completing the square, or using the quadratic formula. This equation does not factor easily, so we'll use the quadratic formula: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \] For the given equation, \(a = 2\), \(b = -4\), and \(c = 4\). Plugging these values into the quadratic formula: \[ x = \frac{{-(-4) \pm \sqrt{{(-4)^2 - 4(2)(4)}}}}{{2(2)}} \] \[ x = \frac{{4 \pm \sqrt{{16 - 32}}}}{{4}} \] \[ x = \frac{{4 \pm \sqrt{{-16}}}}{{4}} \] Since we have a negative number under the square root, the solutions will be complex numbers. We continue: \[ x = \frac{{4 \pm 4i}}{{4}} \] Simplifying further, we divide both terms in the numerator by the denominator: \[ x = 1 \pm i \] The solutions to the quadratic equation are: \[ x = 1 + i \quad \text{and} \quad x = 1 - i \] Here, \(i\) is the imaginary unit, where \(i^2 = -1\).
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