Question

7.4-7.9):2) x2 - 12x + 11 = 0;​

Answer 1

Ответ:

х=11,х=1

рушение обычного квадратного уравнения через теорему Виета

либо дискриминант

Answer 2

It looks like you've provided a quadratic equation: $2x^2 - 12x + 11 = 0$. You're probably looking for the solutions to this equation. To solve it, you can use the quadratic formula:

The quadratic formula states that for an equation $ax^2 + bx + c = 0$, the solutions for $x$ are given by:

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

In your case, $a = 2$, $b = -12$, and $c = 11$. Plugging these values into the quadratic formula:

$x = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 2 \cdot 11}}{2 \cdot 2}$ $x = \frac{12 \pm \sqrt{144 - 88}}{4}$ $x = \frac{12 \pm \sqrt{56}}{4}$ $x = \frac{12 \pm 2\sqrt{14}}{4}$ $x = \frac{6 \pm \sqrt{14}}{2}$

So, the solutions for the equation $2x^2 - 12x + 11 = 0$ are: $x = \frac{6 + \sqrt{14}}{2} \approx 5.37$ $x = \frac{6 - \sqrt{14}}{2} \approx 0.63$

These are the approximate solutions to the quadratic equation.