Question

X²-3x-4=02x²-8=06x²-12x​

Answer 1

1) Д=9-4*1*(-4)=25

х1=(3+5)/2=4

х2=(3-5)/2=-1

2) 2х^2=8

х^2=4

х1=2

х2=-2

3) 6х(х-2)=0

6х=0

х=0

или х-2=0

х=2

4) Д=4+4*3=16

х1=(2+4)/2=3

х2=(2-4)/2=-1

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(-беск, - 1), (3,+беск)

Answer 2

Let's solve each equation step by step:

1. $x^2 - 3x - 4 = 0$:

To solve this quadratic equation, we can use the quadratic formula:

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

where the equation is in the form $ax^2 + bx + c = 0$.

Comparing the equation $x^2 - 3x - 4 = 0$ with the general form, we have $a = 1$, $b = -3$, and $c = -4$.

Substituting these values into the quadratic formula:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)}$,

$x = \frac{3 \pm \sqrt{9 + 16}}{2}$,

$x = \frac{3 \pm \sqrt{25}}{2}$,

$x = \frac{3 \pm 5}{2}$.

Simplifying further, we have two possible solutions:

$x_1 = \frac{3 + 5}{2} = 4$,

$x_2 = \frac{3 - 5}{2} = -1$.

Therefore, the solutions to the equation $x^2 - 3x - 4 = 0$ are $x = 4$ and $x = -1$.

2. $2x^2 - 8 = 0$:

Dividing both sides of the equation by 2, we have:

$x^2 - 4 = 0$.

This equation is a difference of squares, so we can factor it as:

$(x - 2)(x + 2) = 0$.

Setting each factor equal to zero, we get:

$x - 2 = 0$ or $x + 2 = 0$.

Solving for $x$ in each case, we have:

$x = 2$ or $x = -2$.

Therefore, the solutions to the equation $2x^2 - 8 = 0$ are $x = 2$ and $x = -2$.

3. $6x^2 - 12x = 0$:

Factoring out the common factor of $6x$, we have:

$6x(x - 2) = 0$.

Setting each factor equal to zero, we get:

$6x = 0$ or $x - 2 = 0$.

Solving for $x$ in each case, we have:

$x = 0$ or $x = 2$.

Therefore, the solutions to the equation $6x^2 - 12x = 0$ are $x = 0$ and $x = 2$.