Question

Решите уравнение 1) 28х2-36х+11=0 2) -49х2=+21х-2=0 Помогите пожалуйста

Answer 1

$\displaystyle 28x^2-36x+11=0\\ D/4=18^2-11\cdot28=4\cdot(9^2-11\cdot7)=4\cdot(81-77)=4^2\\ x=\frac{18\pm4}{28}\\ x_1=\frac{14}{28}=\frac12;\quad x_2=\frac{22}{28}=\frac{11}{14}$

$-49x^2+21x-2=0\\ (7x)^2-3\cdot7x+2=0\\ 7x=t:~t^2-3t+2=0\\ D=3^2-4\cdot1\cdot2=1\\ t=\dfrac{3\pm1}2\\ t_1=1;\quad t_2=2\\ 7x_1=1;\quad 7x_2=2\\ x_1=\dfrac17;\quad x_2=\dfrac27$

Answer 2

Solving the equation 1: 28x^2 - 36x + 11 = 0

To solve the quadratic equation 28x^2 - 36x + 11 = 0, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = 28, b = -36, and c = 11. Plugging these values into the quadratic formula, we get:

x = (-(-36) ± √((-36)^2 - 4 * 28 * 11)) / (2 * 28)

Simplifying further:

x = (36 ± √(1296 - 1232)) / 56

x = (36 ± √64) / 56

Now, let's calculate the solutions:

x1 = (36 + 8) / 56 = 44 / 56 = 11 / 14

x2 = (36 - 8) / 56 = 28 / 56 = 1 / 2

Therefore, the solutions to the equation 28x^2 - 36x + 11 = 0 are x = 11/14 and x = 1/2.

Solving the equation 2: -49x^2 + 21x - 2 = 0

To solve the quadratic equation -49x^2 + 21x - 2 = 0, we can again use the quadratic formula. Comparing this equation to the general form ax^2 + bx + c = 0, we have a = -49, b = 21, and c = -2. Plugging these values into the quadratic formula, we get:

x = (-21 ± √(21^2 - 4 * (-49) * (-2))) / (2 * (-49))

Simplifying further:

x = (-21 ± √(441 - 392)) / (-98)

x = (-21 ± √49) / (-98)

Now, let's calculate the solutions:

x1 = (-21 + 7) / (-98) = -14 / (-98) = 1 / 7

x2 = (-21 - 7) / (-98) = -28 / (-98) = 2 / 7

Therefore, the solutions to the equation -49x^2 + 21x - 2 = 0 are x = 1/7 and x = 2/7.

I hope this helps! Let me know if you have any further questions.