Question

Допоможіть з прикладами які обведені зеленим

Answer 1

$a {x}^{2} + bx + c = a(x - x_{1})(x - x_{2})$

1.

2)

$a = 1 \\ x _{1} = - 2 \\ x_{2} = 3 \\ (x - ( - 2))(x - 3) = (x +2 )(x - 3) = \\ {x}^{2} - 3x + 2x - 6 = {x}^{2} - x - 6$

3)

$a = 5\\ x _{1} = \frac{2}{5} \\ x_{2} = - 5 \\ 5(x - \frac{2}{5} )(x - ( - 5)) = (5x - 2)(x + 5) = \\ 5 {x}^{2} + 25x -2 x - 10 = 5 {x}^{2} + 23x - 10$

2.

1)

По теореме Виета:

${x}^{2} + bx + c = 0 \\ x _{1} + x_{2} = - b \\ x _{1} x_{2} = c$

${x}^{2} + 3x + 2 = 0 \\ x _{1} + x_{2} = - 3 \\ x _{1} x_{2} = 2 \\ x _{1} = - 1\\ x_{2} = - 2 \\ {x}^{2} + 3x + 2 = (x - ( - 1))(x - ( - 2)) = \\ (x + 1)(x + 2)$

4)

${x}^{2} + 6x + 9 = {x}^{2} + 2 \times x \times 3 + {3}^{2} = \\ (x + 3) {}^{2}$

3.

1)

$2 {x}^{2} - 5x + 3 = 0 \\ a = 2 \\ b = - 5 \\ c = 3\\ d = {b }^{2} - 4ac = ( - 5) {}^{2} - 4 \times 2 \times 3 = \\ 25 - 2 = 1 \\ x _{1} = \frac{5 - 1}{2 \times 2} = \frac{4}{4} = 1\\ x_{2} = \frac{5 + 1}{2 \times 2} = \frac{6}{4} = \frac{3}{2} = 1.5 \\ \\ 2 {x}^{2} - 5x + 3 = 2(x - 1)(x - 1.5) = \\ (x - 1)(2x - 3)$

4)

$6 {x}^{2} - 5x + 1 = 0\\ a = 6\\ b = - 5\\ c = 1 \\ d = {b}^{2} - 4ac = ( - 5) {}^{2} - 4 \times 6 \times 1 = \\ 25 - 24 = 1 \\ x _{1} = \frac{5 - 1}{2 \times 6} = \frac{4}{12} = \frac{1}{3} \\ x_{2} = \frac{5 + 1}{2 \times 6} = \frac{6}{12} = \frac{1}{2} \\ 6 {x}^{2} - 5x + 1 = 6(x - \frac{1}{3} )(x - \frac{1}{2} ) = \\ 3(x - \frac{1}{3} ) \times 2(x - \frac{1}{2} ) = \\ (3x - 1)(2x - 1)$

8)

$- 2 {x}^{2} + 3x + 2 = 0 \\ 2 {x}^{2} - 3x - 2 = 0 \\ a = 2 \\ b = - 3 \\ c = - 2\\ d = {b}^{2} - 4ac = ( - 3) {}^{2} - 4 \times 2 \times ( - 2) = \\ 9 + 16 = 25 \\ x _{1} = \frac{3 + 5}{2 \times 2} = \frac{8}{4} = 2\\ x_{2} = \frac{3 - 5}{2 \times 2} = - \frac{2}{4} = - \frac{1}{2} \\ \\ - 2 {x}^{2} + 3x + 2 = - 2(x - 2)(x - ( - \frac{1}{2})) = \\ - 2(x + \frac{1}{2} )(x - 2) = - ( 2x + 1)(x - 2)$