Question

1.(6x+3)(9-x)=02.(x+5)(2x-1/3)=0

Answer 1

Чтобы произведение равнялось 0, достаточно чтобы один из множителей был равен 0.

1. 6х + 3 = 0

6х = - 3

х = - 3 : 6

х = - 3/6 = - 1/2 - первый корень уравнения

9 - х = 0

х = 9 - 0

х = 9 - второй корень уравнения

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

2. х + 5 = 0

х = - 5 - первый корень уравнения

2х - 1/3 = 0

2х = 1/3

х = 1/3 : 2

х = 1/3 * 1/2

х = 1/6 - второй корень уравнения

Answer 2

To find the values of 'x' that satisfy the given equations, we need to set each equation to zero and then solve for 'x' separately. Let's do that:

1. (6x+3)(9-x) = 0

First, factor the expression on the left side: (6x + 3)(9 - x) = 0 3(2x + 1)(9 - x) = 0

Now, set each factor to zero and solve for 'x':

1. 3 = 0 (This is not possible, so we ignore it.)

2. 2x + 1 = 0 2x = -1 x = -1/2

3. 9 - x = 0 x = 9

The values of 'x' that satisfy the equation are x = -1/2 and x = 9.

2. (x + 5)(2x - 1/3) = 0

First, let's expand the expression on the left side: (x + 5)(2x - 1/3) = 0 2x^2 - (1/3)x + 10x - (1/3) = 0 2x^2 + (29/3)x - (1/3) = 0

Now, let's use the quadratic formula to solve for 'x': The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for 'x' are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

For our equation, a = 2, b = 29/3, and c = -1/3.

x = (-(29/3) ± √((29/3)^2 - 4(2)(-1/3))) / 2(2) x = (-(29/3) ± √((841/9) + 8/3)) / 4 x = (-(29/3) ± √(841/9 + 24/9)) / 4 x = (-(29/3) ± √(865/9)) / 4 x = (-(29/3) ± √865/3) / 4

Now we have two possible solutions for 'x':

1. x = (-(29/3) + √865/3) / 4
2. x = (-(29/3) - √865/3) / 4

These are the two values of 'x' that satisfy the equation.