Question

(x+1)(6x2-5x+1)=0 через дискриминант

Answer 1

Думаю сам разберешся

Answer 2

(x+1)(6x 2-5x+1)=0

x+1=0    или    6x 2-5x+1=0

известные переносим в одну сторону, а не известный в другую.

x= -1       или      7x= -1

                          x= 7: (-1)

                          x= -7

ответ: -1, -7.

Answer 3

To find the solutions for the quadratic equation $(x+1)(6x^2-5x+1) = 0$ using the discriminant, we first need to recognize that the given equation is a product of two factors set equal to zero. Therefore, we have two separate equations:

1. $x + 1 = 0$
2. $6x^2 - 5x + 1 = 0$

Let's solve each equation one by one:

1. $x + 1 = 0$

Subtract 1 from both sides: $x = -1$

Now, for the second equation:

2. $6x^2 - 5x + 1 = 0$

The quadratic equation $ax^2 + bx + c = 0$ has solutions given by the quadratic formula:

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

In this case, $a = 6$, $b = -5$, and $c = 1$.

Calculate the discriminant $D$ first:

$D = b^2 - 4ac$ $D = (-5)^2 - 4 \cdot 6 \cdot 1$ $D = 25 - 24$ $D = 1$

Now, we can find the solutions for $x$ using the quadratic formula:

$x = \frac{-(-5) \pm \sqrt{1}}{2 \cdot 6}$ $x = \frac{5 \pm 1}{12}$

So, we get two solutions:

1. $x = \frac{5 + 1}{12} = \frac{6}{12} = \frac{1}{2}$
2. $x = \frac{5 - 1}{12} = \frac{4}{12} = \frac{1}{3}$

Therefore, the solutions to the equation $(x+1)(6x^2-5x+1) = 0$ are $x = -1$, $x = \frac{1}{2}$, and $x = \frac{1}{3}$.