Question

(4x−15)2−4(4x−15)+3=0

Answer 1

Здравствуйте, Dumenko200514!

Если перед скобкой стоит общий множитель, то каждый член скобки умножается на этот множитель.

Решение данного уравнения:

$\tt \Big(4x-15\Big)\cdot2-4\cdot\Big(4x-15\Big)+3=0 \\ \\ \tt 8x-30-16x+60+3=0 \\ \\ \tt -8x+33=0 \\ \\ \tt -8x=-33 \\ \\ \tt x=-33\div\Big(-8\Big) \\ \\ \bf x=4,125$

Окончательный ответ:

Ответ данного уравнения - "x=4,125".

С Уважением, NeNs07.

Answer 2

To solve the quadratic equation (4x−15)²−4(4x−15)+3=0, you can use a substitution to make it easier to work with. Let's set y = 4x - 15, so the equation becomes:

y² - 4y + 3 = 0

Now, we have a standard quadratic equation in the form of ay² + by + c = 0, where a = 1, b = -4, and c = 3. To solve for y, we can use the quadratic formula:

y = (-b ± √(b² - 4ac)) / 2a

Substitute the values into the formula:

y = (4 ± √((-4)² - 4 * 1 * 3)) / 2 * 1 y = (4 ± √(16 - 12)) / 2 y = (4 ± √4) / 2 y = (4 ± 2) / 2

Now, we have two possible solutions for y:

1. y = (4 + 2) / 2 y = 6 / 2 y = 3

2. y = (4 - 2) / 2 y = 2 / 2 y = 1

Now, we need to solve for x by reversing the substitution:

1. When y = 3: 4x - 15 = 3 4x = 3 + 15 4x = 18 x = 18 / 4 x = 4.5

2. When y = 1: 4x - 15 = 1 4x = 1 + 15 4x = 16 x = 16 / 4 x = 4

So, the solutions for the original equation are x = 4.5 and x = 4.