Question

X²-6x+6=0 через Дискриминант​

Answer 1

$x1 = 3 - \sqrt{3 } \\ x2 = 3 + \sqrt{3}$

Ответ:

Answer 2

To solve the quadratic equation $x^2 - 6x + 6 = 0$ using the discriminant, you can follow these steps:

The quadratic equation is in the form $ax^2 + bx + c = 0$, where $a = 1$, $b = -6$, and $c = 6$.

The discriminant is given by $D = b^2 - 4ac$.

1. Calculate the discriminant: $D = (-6)^2 - 4 \cdot 1 \cdot 6$ $D = 36 - 24$ $D = 12$

2. Determine the nature of the roots based on the discriminant value:

   • If $D > 0$, the quadratic equation has two distinct real roots.
   • If $D = 0$, the quadratic equation has one real root (a repeated root).
   • If $D < 0$, the quadratic equation has no real roots (two complex conjugate roots).

In our case, since $D = 12 > 0$, the quadratic equation has two distinct real roots.

3. Use the quadratic formula to find the roots: The quadratic formula is given by $x = \frac{-b \pm \sqrt{D}}{2a}$.

   $x = \frac{-(-6) \pm \sqrt{12}}{2 \cdot 1}$ $x = \frac{6 \pm \sqrt{12}}{2}$ $x = \frac{6 \pm 2\sqrt{3}}{2}$

Now, you have two solutions for $x$:

$x_1 = \frac{6 + 2\sqrt{3}}{2} = 3 + \sqrt{3} \approx 4.732$

$x_2 = \frac{6 - 2\sqrt{3}}{2} = 3 - \sqrt{3} \approx 1.268$

So the two distinct real roots of the quadratic equation $x^2 - 6x + 6 = 0$ are approximately $4.732$ and $1.268$.