Question

3) (x-6) (x+6) - (2x - 3)(x-1)=6-22.​

Answer 1

Ответ:

Объяснение:

x²-36-2x²+2x+3x-3= -16

-x²+5x-23=0

x²-5x+23=0

D=25-92= -67

D<0 решений нет

Answer 2

Объяснение:

х²-36-2х²+2х+3х-3=-16

-х²+5х-39+16=0

-х²+5х-23=0 |÷(-1)

х²-5х+23=0

D=b²-4ac=25-4×23×1=25-92=-67

D<0

ответ: ∅ (пустое множество)

Answer 3

To solve the equation, first, simplify the left-hand side of the equation:

(x - 6)(x + 6) - (2x - 3)(x - 1) = 6 - 22

Expand the expressions:

(x^2 + 6x - 6x - 36) - (2x^2 - 2x - 3x + 3) = -16

Combine like terms:

x^2 - 36 - (2x^2 - 5x + 3) = -16

Distribute the negative sign:

x^2 - 36 - 2x^2 + 5x - 3 = -16

Combine like terms again:

-x^2 + 5x - 39 = -16

Now, move all terms to one side of the equation to set it equal to zero:

-x^2 + 5x - 39 + 16 = 0

Simplify:

-x^2 + 5x - 23 = 0

Next, let's solve this quadratic equation. We can use the quadratic formula:

The quadratic formula is given by:

x = (-b ± √(b^2 - 4ac)) / 2a

In this equation, a = -1, b = 5, and c = -23. Plug these values into the formula:

x = (-(5) ± √(5^2 - 4(-1)(-23))) / 2(-1) x = (-5 ± √(25 - 92)) / -2 x = (-5 ± √(-67)) / -2

Since the expression inside the square root is negative (-67), the quadratic equation has no real solutions. This means the original equation does not have any real solutions. If you are looking for complex solutions, they can be expressed as:

x = (-5 ± i√67) / -2

where "i" represents the imaginary unit (√(-1)).