Question

(a^2-a-2)*x^2+(a+1)*x+1=0

Answer 1

Ответ:

(a^2-a-2)*x^2+(a+1)*x+1=0

$D=(a+1)^2-4(a^2-a-2)*1$

$></p> <p><img src=$

Рассмотрим все возможные случаи

$\left \{ {{-3a^2+6a+9>0} \atop {-3a^2+6a+9=0}} \atop {-3a^2+6a+9$

$-3a^2+6a+9>0$

$a^2-2a-3$

$a^2+a-3a-3$

$a(a+1)-3(a+1)$

$(a+1)*(a-3)$

$\left \{ {{a+10}} \right.$

$\left \{ {{a+1>0} \atop {a-3$

$\left \{ {{a3}} \right.$

$\left \{ {{a>-1} \atop {a$

a∈$(-1;3)$

$-3a^2+6a+9=0$

$a^2-2a-3=0$

$a^2+a-3a-3=0$

$a(a+1)-3(a+1)=0$

$(a+1)(a-3)=0$

$a+1=0$

$a-3=0$

$a=-1$

$a=3$

$-3a^2+6a+9$

$a^2-2a-3>0$

$a^2+a-3a-3>0$

$a(a+1)-3(a+1)>0$

$(a+1)(a-3)>0$

$\left \{ {{a+1>0} \atop {a-3>0}} \right.$

$\left \{ {{a+1$

$\left \{ {{a>-1} \atop {a>3}} \right.$

$\left \{ {{a$

a∈(3;+∞)

a∈(-∞;-1)

a∈(-1;2)∪(2;3), 2 действительных корня

a=-1,a=3, 1 действительный корень

a∈(-∞;-1)∪(3;+∞), нет действительных корней

Answer 2

The given equation is a quadratic equation in terms of variable $x$ with a parameter $a$:

$(a^2 - a - 2) \cdot x^2 + (a + 1) \cdot x + 1 = 0$

To solve for $x$, we can use the quadratic formula:

For a quadratic equation $ax^2 + bx + c = 0$, the solutions for $x$ are given by:

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

In this case, the coefficients are: $a = a^2 - a - 2$ $b = a + 1$ $c = 1$

Plugging these values into the quadratic formula, we get:

$x = \frac{-(a + 1) \pm \sqrt{(a + 1)^2 - 4 \cdot (a^2 - a - 2) \cdot 1}}{2 \cdot (a^2 - a - 2)}$

Simplifying under the square root:

$(a + 1)^2 - 4 \cdot (a^2 - a - 2) = a^2 + 2a + 1 - 4a^2 + 4a + 8 = -3a^2 + 6a + 9$

Substituting this back into the formula:

$x = \frac{-(a + 1) \pm \sqrt{-3a^2 + 6a + 9}}{2 \cdot (a^2 - a - 2)}$

Now, you have the solutions for $x$ in terms of the parameter $a$.