Question

formati ecuația de gradul al doilea care are rădăcinile: a) 1-i√3/4. și 1+i√3/4. b) 1-i/1+i si 1+i/1-i urgent!!​

Answer 1

Explicație pas cu pas:

a)

$a + b = \frac{1-i√3 }{4} + \frac{1+i√3}{4} = \frac{1-i√3 + 1+i√3}{4} = \frac{2}{4} = \frac{1}{2}$

$a \times b =( \frac{1-i√3}{4}) (\frac{1+i√3}{4} ) = \frac{1 + 3}{16} = \frac{4}{16} = \frac{1}{4}$

${x}^{2} - (a + b)x + ab = 0 \\ {x}^{2} - \frac{1}{2} x + \frac{1}{4} = 0 \\ 4 {x}^{2} - 2x + 1 = 0$

b)

$a + b = \frac{1-i }{1 + i} + \frac{1 + i}{1-i} = \frac{ {(1 - i)}^{2} + {(1 + i)}^{2} }{{(1 + i)}{(1 - i)}} = \frac{1 - 2i + {i}^{2} + 1 + 2i + {i}^{2}}{1 - {i}^{2} } = \frac{2 - 0}{2} = 0$

$a \times b = (\frac{1-i }{1 + i} ) (\frac{1 + i}{1-i} ) = \frac{ {(1 - i)}{(1 + i)}}{{(1 + i)}{(1 - i)}} = \frac{1 - {i}^{2}}{1 - {i}^{2} } = \frac{2}{2} = 1$

${x}^{2} - (a + b)x + ab = 0 \\ {x}^{2} - 0 \times x + 1 = 0 \\ {x}^{2} + 1 = 0$