Question

Решить уравнения 1) x^2-9x+14=02) x^2+2x-15=03) x^2-4x+4=04) x^2-19x+90=05) x^2-6x+20=0​

Answer 1

1)x² + 9x + 14 = 0

D= 81 - 4 • 4 = 25

x(1)= (-9 + 5)/2 = -2

x(2)= (-9 - 5)/2 = -7

-7 < -2

Ответ: -2

2) D= b² - 4 ac

x= (-b +/- sqrt(D))/2a

D= 4 + 4 • 15 = 64

x1= (-2 + 8)/2 = 3

x2= (-2 - 8)/2 = -5

3) x² - 4x + 4 = 0

D= b² - 4 ac

D= 16 - 4 • 1 • 4 = 16 - 16 = 0

x1=x2=-b/2=2

x1=x2=2

2 последних примера не могу решить

Answer 2

To solve quadratic equations of the form ax^2 + bx + c = 0, you can use the quadratic formula:

For the equation ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

Let's solve each of the equations you provided:

1. x^2 - 9x + 14 = 0 a = 1, b = -9, c = 14

x = (9 ± √((-9)^2 - 4 * 1 * 14)) / 2 * 1 x = (9 ± √(81 - 56)) / 2 x = (9 ± √25) / 2 x = (9 ± 5) / 2

So the solutions are: x = (9 + 5) / 2 = 14 / 2 = 7 x = (9 - 5) / 2 = 4 / 2 = 2

2. x^2 + 2x - 15 = 0 a = 1, b = 2, c = -15

x = (-2 ± √(2^2 - 4 * 1 * (-15))) / 2 * 1 x = (-2 ± √(4 + 60)) / 2 x = (-2 ± √64) / 2 x = (-2 ± 8) / 2

So the solutions are: x = (-2 + 8) / 2 = 6 / 2 = 3 x = (-2 - 8) / 2 = -10 / 2 = -5

3. x^2 - 4x + 4 = 0 a = 1, b = -4, c = 4

x = (4 ± √((-4)^2 - 4 * 1 * 4)) / 2 * 1 x = (4 ± √(16 - 16)) / 2 x = (4 ± √0) / 2 x = (4 ± 0) / 2

In this case, both solutions coincide: x = 4 / 2 = 2

4. x^2 - 19x + 90 = 0 a = 1, b = -19, c = 90

x = (19 ± √((-19)^2 - 4 * 1 * 90)) / 2 * 1 x = (19 ± √(361 - 360)) / 2 x = (19 ± √1) / 2 x = (19 ± 1) / 2

So the solutions are: x = (19 + 1) / 2 = 20 / 2 = 10 x = (19 - 1) / 2 = 18 / 2 = 9

5. x^2 - 6x + 20 = 0 a = 1, b = -6, c = 20

x = (6 ± √((-6)^2 - 4 * 1 * 20)) / 2 * 1 x = (6 ± √(36 - 80)) / 2 x = (6 ± √(-44)) / 2

Since the discriminant is negative (b^2 - 4ac = -44), the solutions involve imaginary numbers. The solutions will be in the form:

x = (6 + √44i) / 2 or x = (6 - √44i) / 2 x = (6 + 2√11i) / 2 or x = (6 - 2√11i) / 2 x = 3 + √11i or x = 3 - √11i

Note: "i" denotes the imaginary unit, where i^2 = -1.