Question

X^2- 12х + 7 = 0; -t^2+ 5t - 9 = 0; 1/2y^2-2y+2=0 3x^2-11x-1=0 Решите пожалуйста

Answer 1

Ответ:

ответ в закреплённом фото

Answer 2

Quadratic Equation 1: x^2 + 12x + 7 = 0

To solve the quadratic equation x^2 + 12x + 7 = 0, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Comparing the equation to the standard form ax^2 + bx + c = 0, we can see that a = 1, b = 12, and c = 7.

Substituting these values into the quadratic formula, we get:

x = (-12 ± √(12^2 - 4 * 1 * 7)) / (2 * 1)

Simplifying further:

x = (-12 ± √(144 - 28)) / 2

x = (-12 ± √116) / 2

x = (-12 ± √(4 * 29)) / 2

x = (-12 ± 2√29) / 2

Simplifying the expression:

x = -6 ± √29

Therefore, the solutions to the equation x^2 + 12x + 7 = 0 are:

x = -6 + √29 and x = -6 - √29.

Quadratic Equation 2: t^2 + 5t - 9 = 0

To solve the quadratic equation t^2 + 5t - 9 = 0, we can again use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Comparing the equation to the standard form at^2 + bt + c = 0, we can see that a = 1, b = 5, and c = -9.

Substituting these values into the quadratic formula, we get:

t = (-5 ± √(5^2 - 4 * 1 * -9)) / (2 * 1)

Simplifying further:

t = (-5 ± √(25 + 36)) / 2

t = (-5 ± √61) / 2

Therefore, the solutions to the equation t^2 + 5t - 9 = 0 are:

t = (-5 + √61) / 2 and t = (-5 - √61) / 2.

Quadratic Equation 3: (1/2)y^2 - 2y + 2 = 0

To solve the quadratic equation (1/2)y^2 - 2y + 2 = 0, we can follow a similar approach.

Multiplying the entire equation by 2 to eliminate the fraction, we get:

y^2 - 4y + 4 = 0

Comparing the equation to the standard form ay^2 + by + c = 0, we can see that a = 1, b = -4, and c = 4.

Using the quadratic formula, we have:

y = (-(-4) ± √((-4)^2 - 4 * 1 * 4)) / (2 * 1)

Simplifying further:

y = (4 ± √(16 - 16)) / 2

y = (4 ± √0) / 2

y = (4 ± 0) / 2

y = 4 / 2

y = 2

Therefore, the solution to the equation (1/2)y^2 - 2y + 2 = 0 is:

y = 2.

Quadratic Equation 4: 3x^2 - 11x - 1 = 0

To solve the quadratic equation 3x^2 - 11x - 1 = 0, we can once again use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Comparing the equation to the standard form ax^2 + bx + c = 0, we can see that a = 3, b = -11, and c = -1.

Substituting these values into the quadratic formula, we get:

x = (-(-11) ± √((-11)^2 - 4 * 3 * -1)) / (2 * 3)

Simplifying further:

x = (11 ± √(121 + 12)) / 6

x = (11 ± √133) / 6

Therefore, the solutions to the equation 3x^2 - 11x - 1 = 0 are:

x = (11 + √133) / 6 and x = (11 - √133) / 6.

Please note that the solutions provided are based on the given equations.