Teros о n смакомате Sond -4,
[A1 x7-19x-60=0
[B]x2-11x-60=0
[CIr'+19x+60=0
[DIx?+11x+60=0
[A1] Квадратное уравнение x^2 - 7x - 60 = 0
[B] Квадратное уравнение x^2 - 11x - 60 = 0
[С] Квадратное уравнение x^2 + 19x + 60 = 0
[D] Квадратное уравнение x^2 + 11x + 60 = 0
Чтобы решить эти уравнения, нам нужно найти их корни. Для этого воспользуемся дискриминантом:
D = b^2 - 4ac
где a, b и c - коэффициенты квадратного уравнения.
[A1] D = (-7)^2 - 4 * 1 * (-60) = 49 + 240 = 289
sqrt(D) = 17
x1 = (7 + 17) / 2 = 12
x2 = (7 - 17) / 2 = -5
Таким образом, корни уравнения x^2 - 7x - 60 = 0: x1 = 12 и x2 = -5.
[B] D = (-11)^2 - 4 * 1 * (-60) = 121 + 240 = 361
sqrt(D) = 19
x1 = (11 + 19) / 2 = 15
x2 = (11 - 19) / 2 = -4
Таким образом, корни уравнения x^2 - 11x - 60 = 0: x1 = 15 и x2 = -4.
[С] D = (19)^2 - 4 * 1 * 60 = 361 - 240 = 121
sqrt(D) = 11
x1 = (-19 + 11) / 2 = -4
x2 = (-19 - 11) / 2 = -15
Таким образом, корни уравнения x^2 + 19x + 60 = 0: x1 = -4 и x2 = -15.
[D] D = (11)^2 - 4 * 1 * 60 = 121 - 240 = -119
Так как D < 0, квадратное уравнение x^2 + 11x + 60 = 0 не имеет действительных корней.
Вывод:
1) Корни уравнения x^2 - 7x - 60 = 0: x1 = 12 и x2 = -5
2) Корни уравнения x^2 - 11x - 60 = 0: x1 = 15 и x2 = -4
3) Корни уравнения x^2 + 19x + 60 = 0: x1 = -4 и x2 = -15
4) Уравнение x^2 + 11x + 60 = 0 не имеет действительных корней.
To find the solutions to each of the given quadratic equations, we can use the quadratic formula:
For an equation of the form ax^2 + bx + c = 0, the quadratic formula gives us the solutions for x as follows:
x = (-b ± √(b^2 - 4ac)) / (2a)
Now let's solve each equation:
[A] x^2 + 7x - 19x - 60 = 0
Rearranging the terms, we have x^2 - 12x - 60 = 0. Comparing this equation with the general form ax^2 + bx + c = 0, we have:
a = 1, b = -12, c = -60.
Substituting these values into the quadratic formula, we get:
x = (-(-12) ± √((-12)^2 - 4(1)(-60))) / (2(1))
x = (12 ± √(144 + 240)) / 2
x = (12 ± √384) / 2
x = (12 ± 8√6) / 2
x = 6 ± 4√6
So the solutions to equation [A] are x = 6 + 4√6 and x = 6 - 4√6.
[B] x^2 - 11x - 60 = 0
Comparing this equation with the general form, we have:
a = 1, b = -11, c = -60.
Applying the quadratic formula, we get:
x = (-(-11) ± √((-11)^2 - 4(1)(-60))) / (2(1))
x = (11 ± √(121 + 240)) / 2
x = (11 ± √361) / 2
x = (11 ± 19) / 2
So the solutions to equation [B] are x = 15 and x = -4.
[CIr' + 19x + 60 = 0
It seems like there might be a typo in this equation. Can you provide the correct equation format?
To solve the given equations step by step:
A1) To solve the equation x^2 - 19x - 60 = 0:
Step 1: Factorize the quadratic equation.
(x - 20)(x + 3) = 0
Step 2: Set each factor equal to zero and solve for x.
x - 20 = 0 or x + 3 = 0
x = 20 or x = -3
So the solutions for A1 are x = 20 and x = -3.
B) To solve the equation x^2 - 11x - 60 = 0:
Step 1: Factorize the quadratic equation.
(x - 15)(x + 4) = 0
Step 2: Set each factor equal to zero and solve for x.
x - 15 = 0 or x + 4 = 0
x = 15 or x = -4
So the solutions for B are x = 15 and x = -4.
C) The equation given is r' + 19x + 60 = 0. However, it is not a quadratic equation since it contains r'. Without further information, it is not possible to solve this equation.
D) To solve the equation x? + 11x + 60 = 0:
It seems like there might be a typographical error in the equation provided as it is not clear what is meant by "x?".
If the equation were x^2 + 11x + 60 = 0:
Step 1: Factorize the quadratic equation.
(x + 5)(x + 12) = 0
Step 2: Set each factor equal to zero and solve for x.
x + 5 = 0 or x + 12 = 0
x = -5 or x = -12
So the solutions for D are x = -5 and x = -12.