Question

T x -4) (12x ^ 2 + 29x - 8) = 0

Answer 1

Пошаговое объяснение:

Это уравнение можно решить с помощью метода факторизации.

У вас есть уравнение:

(t x - 4)(12x^2 + 29x - 8) = 0

Первый множитель (t x - 4) равен нулю, если t x = 4, следовательно, x = 4/t.

Второй множитель (12x^2 + 29x - 8) может быть разложен на два множителя:

(12x^2 + 29x - 8) = (3x - 1)(4x + 8)

Теперь вы можете найти значения x, при которых второй множитель равен нулю:

1) 3x - 1 = 0 => 3x = 1 => x = 1/3

2) 4x + 8 = 0 => 4x = -8 => x = -2

Итак, у вас есть три корня для уравнения:

x = 4/t, x = 1/3 и x = -2.

Answer 2

To solve the given quadratic equation $(12x^2 + 29x - 8)(x - 4) = 0$, we'll use the zero-product property, which states that if the product of two or more factors is equal to zero, then at least one of the factors must be zero. Therefore, we'll set each factor equal to zero and solve for $x$:

1. $12x^2 + 29x - 8 = 0$

We can use the quadratic formula to solve this quadratic equation:

The quadratic formula is given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our equation, $a = 12$, $b = 29$, and $c = -8$.

Substitute these values into the quadratic formula:

$x = \frac{-29 \pm \sqrt{29^2 - 4 \times 12 \times -8}}{2 \times 12}$

$x = \frac{-29 \pm \sqrt{841 + 384}}{24}$

$x = \frac{-29 \pm \sqrt{1225}}{24}$

$x_1 = \frac{-29 + 35}{24} = \frac{6}{24} = \frac{1}{4}$

$x_2 = \frac{-29 - 35}{24} = \frac{-64}{24} = -\frac{8}{3}$

So, the solutions for the quadratic equation $12x^2 + 29x - 8 = 0$ are $x = \frac{1}{4}$ and $x = -\frac{8}{3}$.

2. $x - 4 = 0$

Solving for $x$, we get $x = 4$.

Therefore, the solutions to the given equation $(12x^2 + 29x - 8)(x - 4) = 0$ are $x = \frac{1}{4}$, $x = -\frac{8}{3}$, and $x = 4$.