Question

4 (4х - 12)² - 7 (4х - 12) + 3 = 0Х1 =? Х2 =?​

Answer 1

Ответ:

x₁ = 3,25; x₂ = 4,0625

Пошаговое объяснение:

4 (4х - 12)² - 7 (4х - 12) + 3 = 0

Это уравнение решается через замену переменной:

$4x - 12 = t$

$4t^2 - 7t + 3 = 0$

$D = 49 - 48 = 1$

$t_1 = \frac{7-1}{8} = \frac{6}{8} = \frac{3}{4}$

$t_2 = \frac{7+1}{8} = \frac{8}{8} = 1$

Теперь возвращаемся к нашему уравнению:

$4x - 12 = t_1\\\\ 4x - 12 = \frac{3}{4}\\\\4x = 12\frac{3}{4}\\\\4x = \frac{51}{4}\\\\x_1 = \frac{51}{4\cdot4} = \frac{51}{16} = 3\frac{3}{16} = 3,1875$

$4x - 12 = t_2\\\\ 4x - 12 = 1\\\\4x = 13\\\\x = \frac{13}{4}\\\\x_2 = 3\frac{1}{4} = 3,25$

Answer 2

To solve the quadratic equation $4(4x - 12)^2 - 7(4x - 12) + 3 = 0$, we can use the quadratic formula. The standard form of a quadratic equation is $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants. In this case, $a = 4$, $b = -7$, and $c = 3$.

The quadratic formula is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Let's plug in the values and solve for $x_1$ and $x_2$:

$x_1 = \frac{-(-7) + \sqrt{(-7)^2 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}$

$x_2 = \frac{-(-7) - \sqrt{(-7)^2 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}$

Simplifying further:

$x_1 = \frac{7 + \sqrt{49 - 48}}{8} = \frac{7 + \sqrt{1}}{8} = \frac{7 + 1}{8} = \frac{8}{8} = 1$

$x_2 = \frac{7 - \sqrt{49 - 48}}{8} = \frac{7 - \sqrt{1}}{8} = \frac{7 - 1}{8} = \frac{6}{8} = \frac{3}{4}$

Therefore, the solutions for $x_1$ and $x_2$ are:

$x_1 = 1$

$x_2 = \frac{3}{4}$