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

$x^2 - (2a + 4)x + 8a = 0$
$D = b^2 - 4ac = (2a + 4)^2 - 4 * 1 * 8a = 4a^2 + 16a + 16 - 32a = 4a^2 - 16a + 16 = (2a - 4)^2$
$(2a - 4)^2 = 0$
2a − 4 = 0
2a = 4
a = 2
при a = 2:
$D = (2 * 2 - 4)^2 = (4 - 4)^2 = 0^2 = 0$
$x = \frac{-b + \sqrt{D}}{2a} = \frac{2a + 4 + \sqrt{0}}{2 * 1} = \frac{2a + 4}{2} = \frac{2(a + 2)}{2} = a + 2 = 2 + 2 = 4$
при a ≠ 2:
D > 0
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2a + 4 + \sqrt{(2a - 4)^2}}{2 * 1} = \frac{2a + 4 + 2a - 4}{2} = \frac{4a}{2} = 2a$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2a + 4 - \sqrt{(2a - 4)^2}}{2 * 1} = \frac{2a + 4 - (2a - 4)}{2 * 1} = \frac{2a + 4 - 2a + 4}{2} = \frac{8}{2} = 4$
Ответ:
при a = 2:
x = 4
при a ≠ 2:
x = 2a; x = 4.

$a^2x^2 - 24ax - 25 = 0$
$D = b^2 - 4ac = (-24a)^2 - 4 * a^2 * (-25) = 576a^2 + 100a^2 = 676a^2$
$676a^2 = 0$
$a^2 = 0$
a = 0
при a = 0:
$0^2 * x^2 - 24 * 0 * x - 25 = 0$
−25 ≠ 0 − нет корней
при a ≠ 0:
D > 0
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{24a + \sqrt{676a^2}}{2a^2} = \frac{24a + 26a}{2a^2} = \frac{50a}{2a^2} = \frac{25}{a}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{24a - \sqrt{676^2}}{2a^2} = \frac{24a - 26a}{2a^2} = \frac{-2a}{2a^2} = -\frac{1}{a}$
Ответ:
при a = 0:
нет корней
при a ≠ 0:
$x = \frac{25}{a}; x = -\frac{1}{a}$.

$3(2a - 1)x^2 - 2(a + 1)x + 1 = 0$
2a − 1 = 0
2a = 1
a = 0,5
при a = 0,5: уравнение будет линейным и будет иметь 1 корень:
$3(2 * 0,5 - 1)x^2 - 2(0,5 + 1)x + 1 = 0$
$3(1 - 1)x^2 - (1 + 2)x + 1 = 0$
−3x + 1 = 0
−3x = −1
$x = \frac{1}{3}$
при a ≠ 0,5: уравнение будет квадратным:
$D = b^2 - 4ac = (-2(a + 1))^2 - 4 * 3(2a - 1) * 1 = 4(a^2 + 2a + 1) - 12(2a - 1) = 4a^2 + 8a + 4 - 24a + 12 = 4a^2 - 16a + 16 = (2a - 4)^2$
$(2a - 4)^2 = 0$
2a − 4 = 0
2a = 4
a = 2
при a = 2:
$D = (2 * 2 - 4)^2 = (4 - 4)^2 = 0^2 = 0$
$x = \frac{-b + \sqrt{D}}{2a} = \frac{2(a + 1) + \sqrt{0}}{2 * 3(2a - 1)} = \frac{2(a + 1)}{6(2a - 1)} = \frac{2 * (2 + 1)}{6 * (2 * 2 - 1)} = \frac{2 * 3}{6 * (4 - 1)} = \frac{6}{6 * 3} = \frac{1}{3}$
при a ≠ 2:
D > 0
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2(a + 1) + \sqrt{(2a - 4)^2}}{2 * 3(2a - 1)} = \frac{2a + 2 + 2a - 4}{6(2a - 1)} = \frac{4a - 2}{6(2a - 1)} = \frac{2(2a - 1)}{6(2a - 1)} = \frac{1}{3}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2(a + 1) - \sqrt{(2a - 4)^2}}{2 * 3(2a - 1)} = \frac{2a + 2 - (2a - 4)}{6(2a - 1)} = \frac{2a + 2 - 2a + 4}{6(2a - 1)} = \frac{6}{6(2a - 1)} = \frac{1}{2a - 1}$
Ответ:
при a = 0,5:
$x = \frac{1}{3}$
при a = 2:
$x = \frac{1}{3}$
при a ≠ 2:
$x = \frac{1}{3}; x = \frac{1}{2a - 1}$.