Question

1)-3x^2-15x=0 2)-25x^2+36=0 3)x^2-x56=0 помогите пожалуста

Answer 1

1]
1-3x^2-15x=0
-3x(x+5)=0
x1 = 0
x2 = -5

2]
-25x^2+36=0
-25x^2=-36
25x^2=36
x^2=36/25
x1=6/5
x2=-6/5

3]
x^2-56x=0
x(x-56)=0
x1=0
x2=56

Answer 2

Solving the Equations

Let's solve the given equations step by step:

1) -3x^2 - 15x = 0

To solve this equation, we can factor out the common factor of -3x:

-3x(x + 5) = 0

Now, we have two factors that can equal zero:

-3x = 0 or x + 5 = 0

Solving for x in each equation:

-3x = 0 => x = 0

x + 5 = 0 => x = -5

Therefore, the solutions to the equation -3x^2 - 15x = 0 are x = 0 and x = -5.

2) -25x^2 + 36 = 0

To solve this equation, we can use the factoring method:

-25x^2 + 36 = 0

This equation is a quadratic equation in the form of ax^2 + bx + c = 0, where a = -25, b = 0, and c = 36.

We can factor this equation as a difference of squares:

(-5x + 6)(5x + 6) = 0

Now, we have two factors that can equal zero:

-5x + 6 = 0 or 5x + 6 = 0

Solving for x in each equation:

-5x + 6 = 0 => -5x = -6 => x = 6/5

5x + 6 = 0 => 5x = -6 => x = -6/5

Therefore, the solutions to the equation -25x^2 + 36 = 0 are x = 6/5 and x = -6/5.

3) x^2 - x - 56 = 0

To solve this equation, we can use the factoring method or the quadratic formula. Let's use the factoring method:

x^2 - x - 56 = 0

This equation is a quadratic equation in the form of ax^2 + bx + c = 0, where a = 1, b = -1, and c = -56.

We need to find two numbers that multiply to give -56 and add up to -1. The numbers -8 and 7 satisfy these conditions:

(x - 8)(x + 7) = 0

Now, we have two factors that can equal zero:

x - 8 = 0 or x + 7 = 0

Solving for x in each equation:

x - 8 = 0 => x = 8

x + 7 = 0 => x = -7

Therefore, the solutions to the equation x^2 - x - 56 = 0 are x = 8 and x = -7.

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