Question

(2x²-5x-7)(27-x)=0 Срочноооооо

Answer 1

(2x²-5x-7)(27-x) = 0

(x+1)(x-3.5)(27-x) = 0

Произведение равно нулю, если хотя бы один из множителей равен нулю:

х є {-1; 3.5; 27}

Answer 2

To solve the equation (2x²-5x-7)(27-x) = 0, we need to find the values of x that make the equation true.

To do this, we can use the zero product property, which states that if the product of two factors is equal to zero, then at least one of the factors must be equal to zero.

So, we can set each factor equal to zero and solve for x.

First factor: 2x²-5x-7 = 0 Second factor: 27-x = 0

Let's solve each equation separately:

Solving the first factor: 2x²-5x-7 = 0

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 2, b = -5, and c = -7. Substituting these values into the quadratic formula, we get:

x = (-(-5) ± √((-5)² - 4(2)(-7))) / (2(2))

Simplifying further:

x = (5 ± √(25 + 56)) / 4

x = (5 ± √81) / 4

x = (5 ± 9) / 4

This gives us two possible solutions:

x₁ = (5 + 9) / 4 = 14 / 4 = 3.5

x₂ = (5 - 9) / 4 = -4 / 4 = -1

So, the first factor has two solutions: x = 3.5 and x = -1.

Solving the second factor: 27-x = 0

To solve this equation, we can simply isolate x by subtracting 27 from both sides:

27 - x = 0

-x = -27

Dividing both sides by -1, we get:

x = 27

So, the second factor has one solution: x = 27.

Final Solution:

Therefore, the equation (2x²-5x-7)(27-x) = 0 has three solutions: x = 3.5, x = -1, and x = 27.

Please note that these solutions are obtained by solving the equation mathematically. If you have any further questions or need additional assistance, feel free to ask!