Question

Решить уравнения: 1) x^2-x-42=0 2)x^2+13+12=0 3)x^2+6x+16=0 4)49x^2=0 Помогите пожалуйста, желательно объясните в общих чертах, много пропустил, учитель объяснять не желает...

Answer 1

1) D=1² - 4*(-42)=1+168=169
x₁ = (1-√169)/2=(1-13)/2= -12/2= -6
x₂ = (1+√169)/2 = (1+13)/2=7
Ответ: -6;   7.

2) D=13² - 4*12=169-48=121
x₁ = (-13-√121)/2=(-13-11)/2= -12
x₂ = (-13+√121)/2=(-13+11)/2= -1
Ответ: -12;  -1.

3) D=6² - 4*16 = 36 -64 = -28<0
нет решений.
Ответ: нет решений.

4) 49x² = 0
x=0
Ответ: 0
 
49 -x² =0
x² = 49
x₁ = 7
x₂ = -7 
Ответ: -7;    7.

Answer 2

Solving the Equations

Let's solve the given equations step by step:

1) Equation: x^2 - x - 42 = 0

To solve this quadratic equation, we can factorize it or use the quadratic formula. Let's 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)

For our equation, a = 1, b = -1, and c = -42. Plugging these values into the quadratic formula, we get:

x = (-(-1) ± √((-1)^2 - 4(1)(-42))) / (2(1))

Simplifying further:

x = (1 ± √(1 + 168)) / 2

x = (1 ± √169) / 2

x = (1 ± 13) / 2

So the solutions for this equation are:

x = (1 + 13) / 2 = 14 / 2 = 7

x = (1 - 13) / 2 = -12 / 2 = -6

Therefore, the solutions for the equation x^2 - x - 42 = 0 are x = 7 and x = -6.

2) Equation: x^2 + 13x + 12 = 0

To solve this quadratic equation, we can again use the quadratic formula:

For our equation, a = 1, b = 13, and c = 12. Plugging these values into the quadratic formula, we get:

x = (-13 ± √(13^2 - 4(1)(12))) / (2(1))

Simplifying further:

x = (-13 ± √(169 - 48)) / 2

x = (-13 ± √121) / 2

x = (-13 ± 11) / 2

So the solutions for this equation are:

x = (-13 + 11) / 2 = -2 / 2 = -1

x = (-13 - 11) / 2 = -24 / 2 = -12

Therefore, the solutions for the equation x^2 + 13x + 12 = 0 are x = -1 and x = -12.

3) Equation: x^2 + 6x + 16 = 0

Again, let's use the quadratic formula to solve this equation:

For our equation, a = 1, b = 6, and c = 16. Plugging these values into the quadratic formula, we get:

x = (-6 ± √(6^2 - 4(1)(16))) / (2(1))

Simplifying further:

x = (-6 ± √(36 - 64)) / 2

x = (-6 ± √(-28)) / 2

Since we have a negative value under the square root, this equation does not have real solutions. The solutions would involve complex numbers.

4) Equation: 49x^2 = 0

To solve this equation, we can simply divide both sides by 49:

x^2 = 0 / 49

x^2 = 0

Taking the square root of both sides, we get:

x = ±√0

Since the square root of 0 is 0, the solutions for this equation are:

x = 0

In summary, the solutions for the given equations are:

1) x = 7, -6 2) x = -1, -12 3) This equation does not have real solutions. 4) x = 0

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