Example Question - imaginary numbers
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Solving Equations Involving Imaginary Numbers
To solve the equation \( x^2 = -169 \), you'll need to take the square root of both sides. However, because the right side of the equation is negative, it means that you will be dealing with imaginary numbers, since the square root of a negative number is not defined within the real numbers. Here are the steps: \[ x^2 = -169 \] Take the square root of both sides: \[ \sqrt{x^2} = \sqrt{-169} \] Since \( \sqrt{x^2} = x \) and \( \sqrt{-169} = \sqrt{-1} \cdot \sqrt{169} \), and \( \sqrt{-1} \) is defined as the imaginary unit \( i \), you'll have: \[ x = \pm 13i \] The \( \pm \) symbol indicates that there are two solutions for \( x \), one positive and one negative. So the solutions to the equation are \( x = 13i \) and \( x = -13i \).
Solving a Complex Equation with Imaginary Numbers
The image displays a mathematical problem: \[ \text{Find the value of a and b if } bi^{2} + \frac{4}{3}i - \sqrt{3} = a + bi\sqrt{3}. \] To solve this, we need to realize that \(i^2 = -1\) and group real and imaginary parts separately. \[ bi^{2} + \frac{4}{3}i - \sqrt{3} = a + bi\sqrt{3}, \] \[ b(-1) + \frac{4}{3}i - \sqrt{3} = a + bi\sqrt{3}. \] Now, group real parts together and imaginary parts together: Real parts: \(-b - \sqrt{3} = a,\) Imaginary parts: \(\frac{4}{3}i = bi\sqrt{3}.\) Now let's isolate \(b\) from the imaginary part: \[ b = \frac{\frac{4}{3}}{\sqrt{3}} = \frac{4}{3\sqrt{3}} = \frac{4}{3}\cdot\frac{\sqrt{3}}{3} = \frac{4\sqrt{3}}{9}.\] Then substitute \(b\) into the real part to find \(a\): \[ -\left(\frac{4\sqrt{3}}{9}\right) - \sqrt{3} = a,\] \[ a = -\left(\frac{4\sqrt{3}}{9} + \frac{9\sqrt{3}}{9}\right), \] \[ a = -\left(\frac{4\sqrt{3} + 9\sqrt{3}}{9}\right), \] \[ a = -\left(\frac{13\sqrt{3}}{9}\right). \] Thus, the values for \(a\) and \(b\) are: \[ a = -\frac{13\sqrt{3}}{9}, \] \[ b = \frac{4\sqrt{3}}{9}. \]
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