Question

A) 3x2 - 12=0; 6) 2x2 + 6x=0; B) 1,8x²=0; r) x2 +9=0;

Answer 1

Ответ: а) х^2=4 или х1=2, х2=-2

Б) х*(2*х+6)=0 или х1=0, х2=-3

В) х=0

Г) нет решения.

Пошаговое объяснение:

Answer 2

Пошаговое объяснение:

а) 3x2 - 12=0

6х-12=0

х=12÷6

х=2

d) 2x2+6x=0

4x+6x=0

10x=0

x=0

2

B) 1,8x =0

9

_ *x=0

5

81

_ X = 0

25

x=0

r) x2+9=0

Answer 3

It looks like you've provided a set of quadratic equations, and you want to find the solutions for each equation. Let's solve them one by one:

A) 3x^2 - 12 = 0: First, we can simplify the equation by dividing both sides by 3: x^2 - 4 = 0

Now, let's solve for x by taking the square root of both sides: x^2 = 4

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

So, for equation A, the solutions are x = 2 and x = -2.

6. 2x^2 + 6x = 0: Let's factor out a common term, which is 2x: 2x(x + 3) = 0

Now, we can use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. So we have two possibilities:

1. 2x = 0 Divide both sides by 2: x = 0

2. x + 3 = 0 Subtract 3 from both sides: x = -3

So, for equation 6, the solutions are x = 0 and x = -3.

B) 1.8x² = 0: Divide both sides by 1.8: x^2 = 0

Now, take the square root of both sides: x = 0

So, for equation B, the only solution is x = 0.

r) x^2 + 9 = 0: Subtract 9 from both sides: x^2 = -9

Now, take the square root of both sides. Remember that the square root of a negative number is an imaginary number: x = ±√(-9)

x = ±3i

So, for equation r, the solutions are x = 3i and x = -3i, where "i" represents the imaginary unit.