Question

4) x² = 0; 12x + 27 = 0; 6) x² + 9x22 = 0.​

Answer 1

4) х=0; х=-9/4

6) уравнение не имеет действительных решений

Answer 2

It looks like you've provided three different equations:

1. $x^2 = 0$
2. $12x + 27 = 0$
3. $x^2 + 9x^2 = 0$

Let's solve them one by one:

1. $x^2 = 0$

In this case, the only solution is $x = 0$ because any number squared is equal to 0.

2. $12x + 27 = 0$

Subtracting 27 from both sides gives:

$12x = -27$

Dividing both sides by 12:

$x = -\frac{27}{12} = -\frac{9}{4}$

So, the solution is $x = -\frac{9}{4}$.

3. $x^2 + 9x^2 = 0$

Combining like terms on the left side:

$10x^2 = 0$

Dividing both sides by 10:

$x^2 = 0$

Taking the square root of both sides:

$x = 0$

So, the solution is $x = 0$ in this case as well.