Question

15-5x²=0 27-x²=0 3a²-3a+1=0 С решением)

Answer 1

$5x = 15 \\ x = 15 \div 5 \\ x = 3$

Answer 2

To find the solutions to the given equations, we need to solve for the variables.

1. 15 - 5x² = 0 To solve this quadratic equation, we'll set it equal to zero and then factor it (if possible):

15 - 5x² = 0

Now, we can factor out the common factor of 5:

5(3 - x²) = 0

Now, set each factor equal to zero:

5 = 0 OR 3 - x² = 0

The first equation, 5 = 0, is not possible as 5 cannot be equal to zero. Now, let's solve the second equation:

3 - x² = 0

Subtract 3 from both sides:

• x² = -3

Now, divide both sides by -1 (since dividing by a negative number flips the inequality sign):

x² = 3

Finally, take the square root of both sides:

x = ±√3

So, the solutions to the equation 15 - 5x² = 0 are x = √3 and x = -√3.

2. 27 - x² = 0 To solve this quadratic equation, we'll set it equal to zero and then factor it (if possible):

27 - x² = 0

We can factor this as the difference of squares:

(√27 - x)(√27 + x) = 0

Now, set each factor equal to zero:

√27 - x = 0 OR √27 + x = 0

Solve for x in each equation:

x = √27 OR x = -√27

Now, simplify the square root of 27:

x = √(3 * 3 * 3) OR x = -√(3 * 3 * 3)

x = 3 OR x = -3

So, the solutions to the equation 27 - x² = 0 are x = 3 and x = -3.

3. 3a² - 3a + 1 = 0 To solve this quadratic equation, we'll use the quadratic formula since it doesn't seem easily factorable:

The quadratic formula is given by: a = (-b ± √(b² - 4ac)) / 2a

In this equation, a = 3, b = -3, and c = 1:

a = (-(-3) ± √((-3)² - 4 * 3 * 1)) / 2 * 3 a = (3 ± √(9 - 12)) / 6 a = (3 ± √(-3)) / 6

Since there is a square root of a negative number, the solutions will be complex numbers:

a = (3 + √3i) / 6 OR a = (3 - √3i) / 6

So, the solutions to the equation 3a² - 3a + 1 = 0 are a = (3 + √3i) / 6 and a = (3 - √3i) / 6, where "i" represents the imaginary unit (i.e., √(-1)).