$x^2 - (2a - 5)x - 3a^2 + 5a = 0$
$D = b^2 - 4ac = (-(2a - 5))^2 - 4 * 1 * (-3a^2 + 5a) = 4a^2 - 20a + 25 + 12a^2 - 20a = 16a^2 - 40a + 25 = (4a - 5)^2$
$(4a - 5)^2 = 0$
4a − 5 = 0
4a = 5
a = 1,25
при a = 1,25:
$D = (4 * 1,25 - 5)^2 = (5 - 5)^2 = 0^2 = 0$
$x = \frac{-b + \sqrt{D}}{2a} = \frac{2a - 5 + \sqrt{0}}{2 * 1} = \frac{2a - 5}{2} = \frac{2 * 1,25 - 5}{2} = \frac{2,5 - 5}{2} = \frac{-2,5}{2} = -1,25$
при a ≠ 1,25:
D > 0
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2a - 5 + \sqrt{(4a - 5)^2}}{2 * 1} = \frac{2a - 5 + 4a - 5}{2} = \frac{6a - 10}{2} = \frac{2(3a - 5)}{2} = 3a - 5$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2a - 5 - \sqrt{(4a - 5)^2}}{2 * 1} = \frac{2a - 5 - (4a - 5)}{2} = \frac{2a - 5 - 4a + 5}{2} = \frac{-2a}{2} = -a$
Ответ:
при a = 1,25:
x = −1,25
при a ≠ 1,25:
x = 3a − 5; x = −a.

$x^2 + (3a - 4)x - 12a = 0$
$D = b^2 - 4ac = (3a - 4)^2 - 4 * 1 * (-12a) = 9a^2 - 24a + 16 + 48a = 9a^2 + 24a + 16 = (3a + 4)^2$
$(3a + 4)^2 = 0$
3a + 4 = 0
3a = −4
$a = -\frac{4}{3} = -1\frac{1}{3}$
при $a = -1\frac{1}{3}$:
$D = (3 * (-\frac{4}{3}) + 4)^2 = (-4 + 4)^2 = 0^2 = 0$
$x = \frac{-b + \sqrt{D}}{2a} = \frac{-(3a - 4) + \sqrt{0}}{2 * 1} = \frac{-3a + 4}{2} = \frac{-3 * (-\frac{4}{3}) + 4}{2} = \frac{4 + 4}{2} = \frac{8}{2} = 4$
при $a ≠ -1\frac{1}{3}$:
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-(3a - 4) + \sqrt{(3a + 4)^2}}{2 * 1} = \frac{-3a + 4 + 3a + 4}{2} = \frac{8}{2} = 4$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-(3a - 4) - \sqrt{(3a + 4)^2}}{2 * 1} = \frac{-3a + 4 - (3a + 4)}{2} = \frac{-3a + 4 - 3a - 4}{2} = \frac{-6a}{2} = -3a$
Ответ:
при $a = -1\frac{1}{3}$:
x = 4
при $a ≠ -1\frac{1}{3}$:
x = −3a; x = 4.

$ax^2 - (a + 1)x + 1 = 0$
при a = 0: уравнение будет линейным и будет иметь 1 корень:
$0 * x^2 - (0 + 1)x + 1 = 0$
−x + 1 = 0
−x = −1
x = 1
$D = b^2 - 4ac = (-(a + 1))^2 - 4 * a * 1 = a^2 + 2a + 1 - 4a = a^2 - 2a + 1 = (a - 1)^2$
a − 1 = 0
a = 1
при a = 1:
$D = (1 - 1)^2 = 0^2 = 0$
$x = \frac{-b + \sqrt{D}}{2a} = \frac{a + 1 + \sqrt{0}}{2a} = \frac{a + 1}{2a} = \frac{1 + 1}{2 * 1} = \frac{2}{2} = 1$
при a ≠ 1:
D > 0
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{a + 1 + \sqrt{(a - 1)^2}}{2a} = \frac{a + 1 + a - 1}{2a} = \frac{2a}{2a} =1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{a + 1 - \sqrt{(a - 1)^2}}{2a} = \frac{a + 1 - (a - 1)}{2a} = \frac{a + 1 - a + 1}{2a} = \frac{2}{2a} = \frac{1}{a}$
Ответ:
при a = 0:
x = 1
при a = 1:
x = 1
при a ≠ 1:
$x = 1; x = \frac{1}{a}.$