Question

X^2+1=0 x^2+4=0 x^2-9x=0 x^2=5x

Answer 1

1)x^2=-1
х = пустое множество
2)x^2=-4
x= пустое множество т.к квадрат отрицательным быть не может

3)x^2-9x=0
x(x-9)=0
x=0
x=9
4)x^2-5x=0
x(x-5)=0
x=0
x=5

Answer 2

Let's solve each equation one by one:

1. x^2 + 1 = 0:

This equation does not have any real solutions because the square of a real number is always non-negative. In this case, x^2 is always greater than or equal to 0, so there are no real numbers x that satisfy this equation.

2. x^2 + 4 = 0:

Similar to the previous equation, this equation also does not have any real solutions. The square of a real number is always non-negative, and adding 4 to a non-negative number will always result in a number greater than or equal to 4. Thus, there are no real numbers x that satisfy this equation.

3. x^2 - 9x = 0:

To solve this equation, we can factor out an x:

x(x - 9) = 0

Setting each factor equal to zero, we have:

x = 0 or x - 9 = 0

If x - 9 = 0, we can solve for x:

x = 9

Therefore, the solutions to this equation are x = 0 and x = 9.

4. x^2 = 5x:

To solve this equation, we can rearrange it to:

x^2 - 5x = 0

Again, we can factor out an x:

x(x - 5) = 0

Setting each factor equal to zero, we have:

x = 0 or x - 5 = 0

If x - 5 = 0, we can solve for x:

x = 5

Therefore, the solutions to this equation are x = 0 and x = 5.