Question

1) 16x-x^2=0 2) x^2-16=0 3)x^2+16=0

Answer 1

Ответ:

1) x=0; 16 / 2)x=±4 / 3)x∈∅

Объяснение:

$><strong></strong></p> <p>x∈∅</p> </div> <div class=$ 0 0

Answer 2

Ответ:1) x=0; x=8

2)x1=4; x2=-4

3) x не имеет корней

Объяснение:

Answer 3

Let's solve each equation:

1. 16x - x^2 = 0

To solve this equation, we first rearrange it to get a quadratic equation in the standard form (ax^2 + bx + c = 0):

x^2 - 16x = 0

Now, we factor out the common term x:

x(x - 16) = 0

Now, we can use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. So,

x = 0 or x - 16 = 0

Now, solve for x in each equation:

x = 0 or x = 16

Therefore, the solutions to the equation are x = 0 and x = 16.

2. x^2 - 16 = 0

This is also a quadratic equation in the standard form:

x^2 - 16 = 0

Now, we can factor the difference of squares:

(x + 4)(x - 4) = 0

Now, use the zero-product property:

x + 4 = 0 or x - 4 = 0

Solve for x:

x = -4 or x = 4

Therefore, the solutions to the equation are x = -4 and x = 4.

3. x^2 + 16 = 0

This is again a quadratic equation in standard form:

x^2 + 16 = 0

However, this equation does not have any real solutions. The left side of the equation is always positive or zero, and it can never equal zero for any real value of x. This is because the square of any real number is always non-negative (positive or zero), and adding 16 to a non-negative number can never result in zero.

Therefore, the equation x^2 + 16 = 0 has no real solutions.