Question

(3x²-5x)²-5(3x²-5x)+6=0​

Answer 1

(3x²-5x)²-5(3x²-5x)+6=0

3х² - 5х = а

а² - 5а + 6 = 0

По теореме Виета:

${x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c$

$a_{1} + a_{2} = 5\\ a_{1}a_{2} =6 \\ a_{1} = 2 \\ a_{2} = 3$

1) а = 2

3х² - 5х = 2

3х² - 5х - 2 = 0

$a = 3\\ b = - 5\\ c = - 2 \\ D = {b}^{2} - 4ac = {( - 5})^{2} - 4 \times 3 \times ( - 2) = 25 + 24 = 49 \\ x_{1} = \frac{ 5 - 7}{ 2\times 3} = - \frac{2}{6} = - \frac{1}{3} \\ x_{2} = \frac{ 5 +7 }{2 \times 3} = \frac{12}{6} = 2$

2) а = 3

3х² - 5х = 3

3х² - 5х - 3 = 0

$a = 3\\ b = - 5 \\ c = - 3\\ D = {b}^{2} - 4ac = ( -5 ) {}^{2} - 4 \times 3 \times ( -3 ) = 25 + 36 = 61 \\ x_{3} = \frac{5 - \sqrt{61} }{2 \times 3} = \frac{5 - \sqrt{61} }{6} \\ x_{4} = \frac{5 + \sqrt{61} }{2 \times 3} = \frac{5 + \sqrt{61} }{6}$

Answer 2

Відповідь:

Уравнение имеет четыре корня:

Х1 = 2,14

Х2 = -0,47

Х3 = 2

Х4 = -1/3

Пояснення:

Имеем уравнение:

( 3Х² - 5Х )² - 5( 3Х² - 5Х ) + 6 = 0​

Сделаем замену:

У = 3Х² - 5Х ( 1 )

Получаем:

У² - 5У + 6 = 0​

Решим квадратное уравнение:

Найдем дискриминант:

D = 5² - 4 × 1 × 6 = 25 - 24 = 1

Найдем корни уравнения:

У1 = ( 5 + 1 ) / 2 = 6 / 2 = 3

У2 = ( 5 - 1 ) / 2 = 4 / 2 = 2

Подставим в качестве значения У в уравнение ( 1 ) найденные корни:

1) У1 = 3

3Х² - 5Х = 3

3Х² - 5Х - 3 = 0

Решим квадратное уравнение:

Найдем дискриминант:

D = 5² - 4 × 3 × ( -3 ) = 25 + 36 = 61

Найдем корни уравнения:

Х1 = ( 5 + √61 ) / 6 = 12,81 / 6 = 2,14

Х2 = ( 5 - √61 ) / 6 = -2,81 / 6 = -0,47

2) У2 = 2

3Х² - 5Х = 2

3Х² - 5Х - 2 = 0

Решим квадратное уравнение:

Найдем дискриминант:

D = 5² - 4 × 3 × ( -2 ) = 25 + 24 = 49

Найдем корни уравнения:

Х3 = ( 5 + √49 ) / 6 = 12 / 6 = 2

Х4 = ( 5 - √49 ) / 6 = -2 / 6 = -1/3

Answer 3

To solve the quadratic equation (3x² - 5x)² - 5(3x² - 5x) + 6 = 0, we can use a substitution to simplify the equation. Let y = 3x² - 5x. Then, we have:

y² - 5y + 6 = 0

Now, we have a quadratic equation in terms of y, which is easier to solve. We can factor this equation:

(y - 2)(y - 3) = 0

Now, we can solve for y:

1. y - 2 = 0 or 2. y - 3 = 0

Solving for y in the first equation:

y - 2 = 0 y = 2

Solving for y in the second equation:

y - 3 = 0 y = 3

Now that we have found the values of y, we can substitute back to find the values of x:

For y = 2: 3x² - 5x = 2

For y = 3: 3x² - 5x = 3

Let's solve each equation separately:

1. 3x² - 5x = 2

We can rearrange this equation and set it equal to zero:

3x² - 5x - 2 = 0

Now, we can factor the quadratic equation:

(3x + 1)(x - 2) = 0

Setting each factor equal to zero and solving for x:

3x + 1 = 0 3x = -1 x = -1/3

x - 2 = 0 x = 2

So, for this equation, we have two solutions: x = -1/3 and x = 2.

2. 3x² - 5x = 3

Rearrange and set it equal to zero:

3x² - 5x - 3 = 0

Now, we can factor this quadratic equation:

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

Setting each factor equal to zero and solving for x:

3x + 1 = 0 3x = -1 x = -1/3

x - 3 = 0 x = 3

So, for this equation, we also have two solutions: x = -1/3 and x = 3.

In total, there are four solutions to the original equation:

x = -1/3, x = 2, x = -1/3, and x = 3.