Question

(6y+5)²=(4y-3)(3y+5)-74=​

Answer 1

Ответ:

$-1.5 и -1.70833$

Пошаговое объяснение:

$(6y + 5)² = (4y - 3)(3y + 5) - 74$

$36y² + 60y + 25 = 12y² + 8y - 15y - 75 - 74$

$36y² + 60y + 25 = 12y² - 7y - 149$

$36y² + 60y + 25 - 12y² + 7y + 149 = 0$

$24y² + 67y + 174 = 0$

$D = b² - 4ac = 67² - 4 * 24 * 174 = 67² - 26568 = 1$

$y₁ = (-b + √D) / 2a = (-67 + 1) / 48 = -1.5$

$y₂ = (-b - √D) / 2a = (-67 - 1) / 48 = -1.70833$

Ответ: уравнение (6y + 5)² = (4y - 3)(3y + 5) - 74 имеет два корня: -1.5 и -1.70833 (приблизительно).

Answer 2

Let's solve the equation step by step:

(6y + 5)² = (4y - 3)(3y + 5) - 74

First, expand the left side of the equation by squaring the binomial (6y + 5)²:

(6y + 5)(6y + 5) = (4y - 3)(3y + 5) - 74

Now, use the distributive property to expand both sides of the equation:

36y² + 30y + 30y + 25 = (12y² + 20y - 9y - 15) - 74

Combine like terms on both sides:

36y² + 60y + 25 = 12y² + 20y - 9y - 15 - 74

Now, let's simplify further:

36y² + 60y + 25 = 12y² + 11y - 89

To isolate all the terms on one side of the equation, subtract both sides by their respective terms:

36y² + 60y + 25 - 12y² - 11y + 89 = 0

Combine like terms again:

(36y² - 12y²) + (60y - 11y) + (25 + 89) = 0

Simplify:

24y² + 49y + 114 = 0

Now, you have a quadratic equation in the form of ax² + bx + c = 0, where a = 24, b = 49, and c = 114.

To solve this quadratic equation, you can use the quadratic formula:

y = (-b ± √(b² - 4ac)) / (2a)

Plug in the values:

y = (-49 ± √(49² - 4 * 24 * 114)) / (2 * 24)

Now, calculate the discriminant (the value inside the square root):

Discriminant = 49² - 4 * 24 * 114 = 2401 - 10944 = -8543

Since the discriminant is negative, the quadratic equation has no real solutions. The solutions are complex numbers. You can find them using the square root of a negative number:

y = (-49 ± √(-8543)) / (2 * 24)

y = (-49 ± i√8543) / 48

So, the solutions for y are:

y = (-49 + i√8543) / 48 y = (-49 - i√8543) / 48

These are the complex solutions to the equation.