Question

1) 5x2-10= 0; 2) x2 + 4x = 0;3) 3x2 + 7x + 2 = 0;4) x2 – 8x +16= 0;5) x2+x+ 3 = 0;6) 3x2-X -5=0.​

Answer 1

Отввет, вот фото с решением

Answer 2

Ответ:

1)x=1

2)x=0

3)x=-2/13

4)x=8/3

5)x=-1

6)x=1

Answer 3

Let's solve each of the given quadratic equations step by step:

1. 5x^2 - 10 = 0: First, let's factor out the common factor, which is 5: 5(x^2 - 2) = 0

Now, set each factor equal to zero and solve for x: x^2 - 2 = 0

To solve this equation, we can take the square root of both sides: x^2 = 2

Taking the square root of both sides gives us two possible solutions: x = √2 or x = -√2

2. x^2 + 4x = 0: We can factor out the common factor, which is x: x(x + 4) = 0

Setting each factor equal to zero: x = 0 or x + 4 = 0

Solving for x, we get: x = 0 or x = -4

3. 3x^2 + 7x + 2 = 0: To solve this quadratic equation, we can factorize it: (3x + 1)(x + 2) = 0

Setting each factor equal to zero: 3x + 1 = 0 or x + 2 = 0

Solving for x in each equation: 3x = -1 or x = -2

Dividing both sides of the first equation by 3: x = -1/3 or x = -2

4. x^2 - 8x + 16 = 0: This equation is already in the form of a perfect square: (x - 4)^2 = 0

Taking the square root of both sides: x - 4 = 0

Solving for x: x = 4

5. x^2 + x + 3 = 0: This quadratic equation does not factorize easily. We can use the quadratic formula to solve it: x = (-b ± √(b^2 - 4ac)) / (2a)

For this equation, a = 1, b = 1, and c = 3: x = (-1 ± √(1^2 - 4(1)(3))) / (2(1)) x = (-1 ± √(1 - 12)) / 2 x = (-1 ± √(-11)) / 2

Since the discriminant (√(-11)) is negative, the equation has no real solutions.

6. 3x^2 - x - 5 = 0: To solve this quadratic equation, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)

For this equation, a = 3, b = -1, and c = -5: x = (-(-1) ± √((-1)^2 - 4(3)(-5))) / (2(3)) x = (1 ± √(1 + 60)) / 6 x = (1 ± √61) / 6

Therefore, the solutions are: x = (1 + √61) / 6 or x = (1 - √61) / 6