Question

1) 4x^2 - 21x + 26 = 0 2) 3x^2 - 8x + 4 = 0 3) 12x^2 - 4x - 1 = 0 4) 6x^2 - 23x + 21 = 0 5) 6x^2 + x - 1 = 0 6) 12x^2 + 28x - 15 = 0

Answer 1

Вот и ответ на задание

Answer 2

To solve the quadratic equations you provided, we can use the quadratic formula, which states that for an equation in the form of ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

Now let's apply the quadratic formula to each of the equations:

1. 4x^2 - 21x + 26 = 0: a = 4, b = -21, c = 26 x = (21 ± √((-21)^2 - 4426)) / (2*4) x = (21 ± √(441 - 416)) / 8 x = (21 ± √25) / 8 x = (21 ± 5) / 8

So the solutions are: x1 = (21 + 5) / 8 = 26 / 8 = 3.25 x2 = (21 - 5) / 8 = 16 / 8 = 2

2. 3x^2 - 8x + 4 = 0: a = 3, b = -8, c = 4 x = (8 ± √((-8)^2 - 434)) / (2*3) x = (8 ± √(64 - 48)) / 6 x = (8 ± √16) / 6 x = (8 ± 4) / 6

So the solutions are: x1 = (8 + 4) / 6 = 12 / 6 = 2 x2 = (8 - 4) / 6 = 4 / 6 = 2/3

3. 12x^2 - 4x - 1 = 0: a = 12, b = -4, c = -1 x = (4 ± √((-4)^2 - 412(-1))) / (2*12) x = (4 ± √(16 + 48)) / 24 x = (4 ± √64) / 24 x = (4 ± 8) / 24

So the solutions are: x1 = (4 + 8) / 24 = 12 / 24 = 0.5 x2 = (4 - 8) / 24 = -4 / 24 = -1/6

4. 6x^2 - 23x + 21 = 0: a = 6, b = -23, c = 21 x = (23 ± √((-23)^2 - 4621)) / (2*6) x = (23 ± √(529 - 504)) / 12 x = (23 ± √25) / 12 x = (23 ± 5) / 12

So the solutions are: x1 = (23 + 5) / 12 = 28 / 12 = 7/3 x2 = (23 - 5) / 12 = 18 / 12 = 3/2

5. 6x^2 + x - 1 = 0: a = 6, b = 1, c = -1 x = (-1 ± √(1^2 - 46(-1))) / (2*6) x = (-1 ± √(1 + 24)) / 12 x = (-1 ± √25) / 12 x = (-1 ± 5) / 12

So the solutions are: x1 = (-1 + 5) / 12 = 4 / 12 = 1/3 x2 = (-1 - 5) / 12 = -6 / 12 = -1/2

6. 12x^2 + 28x - 15 = 0: a = 12, b = 28, c = -15 x = (-28 ± √(28^2 - 412(-15))) / (2*12) x = (-28 ± √(784 + 720)) / 24 x = (-28 ± √1504) / 24 x = (-28 ± √(16 * 94)) / 24 x = (-28 ± 4√94) / 24 x = (-7 ± √94) / 6

So the solutions are: x1 = (-7 + √94) / 6 x2 = (-7 - √94) / 6

These are the solutions for each of the given quadratic equations.