Question

Найдите корни уравнений
1)x^2-6x+8=0
2)x^2-12+11=0
3)3y^2-8y+4=0
4)-2y^2+9y-10=0

Answer 1

N°1: x=2 и х=4

N°2: x=1 и х=11

N°3: y=2/3 и y=2

N°4: y=2 и y=2,5

Объяснение:

Формулы для квадратного уравнения ax²+bx+c = 0

D =b²-4ac

$x _{1} = \frac{ - b \pm \sqrt{D} }{2a} \: \: \: \: \: \: \: x _{2} = \frac{ - b \pm \sqrt{D} }{2a}$

Решаем:

$1) \: {x}^{2} - 6x + 8 = 0 \\ D = {6}^{2} - 4 \times 8 = 4 = {2}^{2} \\ x _{1} = \frac{6 + 2}{2} = 4 \: \: \: \: \: \: \: x _{2} = \frac{6 - 2}{2} = 2$

$2) \: {x}^{2} - 12x + 11 = 0 \\ D = {12}^{2} - 4 \times 11 = 100 = {10}^{2} \\ x _{1} = \frac{12 - 10}{2} = 1 \: \: \: \: \: \: x _{2} = \frac{12 + 10}{2} = 11$

$3) \: 3 {y}^{2} - 8y + 4 = 0 \\ D = {8}^{2} - 4 \times 3 \times 4 = 16 = {4}^{2} \\ y _{1} = \frac{8 + 4}{2 \times 3} = \frac{12}{6} = 2 \: \: \: \: \: \: \: \: y _{2} = \frac{8 - 4}{2 \times 3} = \frac{4}{6} = \frac{2}{3}$

$4) \: - 2 {y}^{2} + 9y - 10 = 0 \: | \times ( - 1) \\ 2 {y}^{2} - 9y + 10 = 0 \\ D = {9}^{2} - 4 \times 2 \times 10 = 1 \\ y _{1} = \frac{9 + 1}{2 \times 2} = \frac{10}{4} = 2.5 \: \: \: \: \: \: \: \: \: y _{2} = \frac{9 - 1}{2 \times 2} = \frac{8}{4} = 2$