Question

4(2x−14)2−19(2x−14)+12=0

Answer 1

Ответ:

Объяснение:

ось і все

Answer 2

To solve the given quadratic equation $4(2x-14)^2 - 19(2x-14) + 12 = 0$, we'll follow these steps:

1. Expand and simplify the equation.
2. Substitute $u = 2x - 14$ to simplify the equation further.
3. Solve the resulting quadratic equation in terms of $u$.
4. Substitute back $u$ to find the values of $x$.

Let's go through these steps.

Step 1: Expand and simplify the equation $4(2x-14)^2 - 19(2x-14) + 12 = 0$

Expand: $4(4x^2 - 56x + 196) - 38x + 266 + 12 = 0$

Simplify: $16x^2 - 224x + 784 - 38x + 278 = 0$ $16x^2 - 262x + 1062 = 0$

Step 2: Substitute $u = 2x - 14$ Let $u = 2x - 14$. Then, $u^2 = (2x - 14)^2$.

Substitute $u$ into the quadratic equation: $16 \left(\frac{u + 14}{2}\right)^2 - 262 \left(\frac{u + 14}{2}\right) + 1062 = 0$

Simplify further: $4u^2 - 262u + 2984 = 0$

Step 3: Solve the resulting quadratic equation in terms of $u$ Use the quadratic formula to solve for $u$: $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 4$, $b = -262$, and $c = 2984$. $u = \frac{262 \pm \sqrt{262^2 - 4 \times 4 \times 2984}}{2 \times 4}$

Calculate: $u_1 = \frac{262 + \sqrt{262^2 - 4 \times 4 \times 2984}}{8}$ $u_2 = \frac{262 - \sqrt{262^2 - 4 \times 4 \times 2984}}{8}$

Step 4: Substitute back $u$ to find the values of $x$ Now, substitute $u$ back in terms of $x$: $u_1 = 2x - 14$ $u_2 = 2x - 14$

Solve for $x$: $2x_1 - 14 = u_1$ $2x_2 - 14 = u_2$

$x_1 = \frac{u_1 + 14}{2}$ $x_2 = \frac{u_2 + 14}{2}$

Plug in the values of $u_1$ and $u_2$ to find $x_1$ and $x_2$