Question

Разложить на множители квадратный трёхчлен

Answer 1

надеюсь помогла.считала дискриминантом

Answer 2

ж) $2 {x}^{2} - 5x + 3 = a(x - x_1 )(x - x_2) = 2(x - \frac{3}{2} )(x - 1) = (2x - 3)(x - 1)$

$2 {x}^{2} - 5x + 3 = 0$

$D = {b}^{2} - 4ac = {(-5)}^{2} - 4 \times 2 \times 3 = 25 - 24 = 1$

$x_1 = \frac{ - b + \sqrt{D} }{2a} = \frac{ -(-5) + \sqrt{1} }{2 \times 2} = \frac{ 5 + 1}{4} = \frac{6}{4} = \frac{3}{2}$

$x_2 = \frac{ - b - \sqrt{D} }{2a} = \frac{ -(-5) - \sqrt{1} }{2 \times 2} = \frac{ 5 - 1}{4} = \frac{4}{4} = 1$

з) $5 {y}^{2} + 2y - 3 = a(y - y_1 )(y - y_2) = 5(y - \frac{3}{5} )(y - ( - 1)) = (5y - 3)(y + 1)$

$5 {y}^{2} + 2y - 3 = 0$

$D = {b}^{2} - 4ac = {2}^{2} - 4 \times 5 \times ( - 3) = 4 + 60 = 64$

$y_1 = \frac{ - b + \sqrt{D} }{2a} = \frac{ - 2 + \sqrt{64} }{2 \times 5} = \frac{ - 2 + 8}{10} = \frac{6}{10} = \frac{3}{5}$

$y_2 = \frac{ - b - \sqrt{D} }{2a} = \frac{ - 2 - \sqrt{64} }{2 \times 5} = \frac{ - 2 - 8}{10} = \frac{ - 10}{10} = - 1$

и) $-2 {x}^{2} + 5x + 7 = a(x - x_1 )(x - x_2) = - 2(x - \frac{7}{2} )(x - ( - 1)) = (-2x + 7)(x + 1) = (7- 2x)(x+1)$

$-2 {x}^{2} + 5x + 7 =0$

$D = {b}^{2} - 4ac = {5}^{2} - 4 \times (-2) \times 7 = 25 + 56 = 81$

$x_1 = \frac{ - b + \sqrt{D} }{2a} = \frac{ -5 + \sqrt{81} }{2 \times (-2)} = \frac{ -5 + 9}{-4} = \frac{4}{-4} = - 1$

$x_2 = \frac{ - b - \sqrt{D} }{2a} = \frac{ -5 - \sqrt{81} }{2 \times (-2)} = \frac{ -5 - 9}{-4} = \frac{-14}{-4} = \frac{7}{2}$