Question

3x^2-5x-2=0 9x^2-6x+1=0​

Answer 1

Пошаговое объяснение:

3х^2-5х-2=0

3х^2+х-6х-2=0

Х(3х+1)-2(3х+1)=0

(3х+1)(х-2)=0

3х+1=0 х-2=0

3х= - 1 х=2

Х= - 1/3

Ответ : х1= - 1/3 х2=2

2)

9х^2-6х+1=0

(3х-1)^2=0

3х-1=0

3х=1

Х=1/3

Ответ : х=1/3

Answer 2

Пошаговое объяснение:

Решаем через дискриминант

Answer 3

To solve these quadratic equations, we can use the quadratic formula:

For a quadratic equation of the form ax^2 + bx + c = 0, the solutions are given by:

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

Let's solve each equation one by one.

1. $3x^2 - 5x - 2 = 0$

Here, a = 3, b = -5, and c = -2.

Using the quadratic formula:

$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-2)}}{2(3)}$

$x = \frac{5 \pm \sqrt{25 + 24}}{6}$

$x = \frac{5 \pm \sqrt{49}}{6}$

$x = \frac{5 \pm 7}{6}$

So, we have two possible solutions:

$x_1 = \frac{12}{6} = 2$

$x_2 = \frac{-2}{6} = -\frac{1}{3}$

2. $9x^2 - 6x + 1 = 0$

Here, a = 9, b = -6, and c = 1.

Using the quadratic formula:

$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(9)(1)}}{2(9)}$

$x = \frac{6 \pm \sqrt{36 - 36}}{18}$

$x = \frac{6}{18} = \frac{1}{3}$

So, the second equation has one solution:

$x = \frac{1}{3}$

To summarize:

1. For the first equation $3x^2 - 5x - 2 = 0$, the solutions are $x = 2$ and $x = -\frac{1}{3}$.

2. For the second equation $9x^2 - 6x + 1 = 0$, the solution is $x = \frac{1}{3}$.