Question

Разложите на линейные множители квадратный трехчлен

Answer 1

в решении

Пошаговое объяснение:

$3)3 {y}^{2} - 11y - 20 = 0 \\ d = ( - 11) {}^{2} - 4 \times 3 \times ( - 20) = 121 + 240 = 361 {19}^{2} \\ x1 = \frac{11 - \sqrt{ {19}^{2} } }{2 \times 3} = \frac{11 - 19}{6} = \frac{ - 8}{6} = \frac{ - 4}{3} \\ x2 = \frac{11 + \sqrt{19 {}^{2} } }{2 \times 3} = \frac{11 + 19}{6} = \frac{30}{6} = 5 \\ 3y {}^{2} - 11y - 20 = a(x - x1)(x - x2) = 3(x + \frac{4}{3} )( x - 5) \\$

$12 {x}^{2} + 7x + 1 = 0 \\ d = {7}^{2} - 4 \times 12 = 49 - 48 = 1 \\ {1}^{2} \\ x1 = \frac{ - 7 - \sqrt[ ]{1 {}^{2} } }{2 \times 12} = \frac{ - 8}{24} = \frac{ - 1}{3} \\ x2 = \frac{ - 7 + \sqrt{ {1}^{2} } }{2 \times 12} = \frac{ - 6}{24} = \frac{ - 1}{4} \\ 12 {x}^{2} + 7x + 1 = 12(x + \frac{1}{3} )(x + \frac{1}{4} )$