Question

1) (x-2)^2-1=0 2) x^2-32=0 3)3x^2=15

Answer 1

Объяснение:

1)(х-2) ^2-1=0

х^2-4х+4-1=0

х^2-4х+3=0

Д=16-4•1•3

Д=4

х1=3

х2=1

2)х^2=32

х=+-√32

х=+-4√2

3)3х^2=15

х^2=5

х=+-√5

Answer 2

To solve each of these equations, follow these steps:

1. (x - 2)^2 - 1 = 0

First, expand the square:

(x - 2)(x - 2) - 1 = 0

x^2 - 4x + 4 - 1 = 0

Now, simplify the equation:

x^2 - 4x + 3 = 0

Next, factor the quadratic equation:

(x - 3)(x - 1) = 0

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

x - 3 = 0 or x - 1 = 0

x = 3 or x = 1

So, the solutions are x = 3 and x = 1.

2. x^2 - 32 = 0

This is a simple quadratic equation. To solve it, add 32 to both sides:

x^2 - 32 + 32 = 0 + 32

x^2 = 32

Now, take the square root of both sides (remembering that there are both positive and negative square roots):

x = ±√32

You can simplify √32 as 4√2:

x = ±4√2

So, the solutions are x = 4√2 and x = -4√2.

3. 3x^2 = 15

First, divide both sides of the equation by 3 to isolate x^2:

x^2 = 15 / 3

x^2 = 5

Now, take the square root of both sides:

x = ±√5

So, the solutions are x = √5 and x = -√5.