Question

Решите уравнение x^2-3x+2=0

Answer 1

$x^{2} -3x+2=0\\D=b^{2} -4ac=9-4*1*2=1\\x_{1} =\frac{-b+\sqrt{D} }{2a} =\frac{3+1}{2} =2\\x_{2} =\frac{-b-\sqrt{D} }{2a} =\frac{3-1}{2} =1$

Answer 2

По теореме Виета для уравнений, где старший коеффициент 1:

Х1+Х2=-b

X1*X2=c

X1+X2=3

X1*X2=2

X1=1

X2=2

Ответ: 1;2

Answer 3

To solve the quadratic equation $x^2 - 3x + 2 = 0$, we can use the quadratic formula:

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

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

In the given equation, $a = 1$, $b = -3$, and $c = 2$. Now, let's plug these values into the formula and solve for $x$:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$ $x = \frac{3 \pm \sqrt{9 - 8}}{2}$ $x = \frac{3 \pm \sqrt{1}}{2}$

Now we have two possible solutions:

1. $x = \frac{3 + 1}{2} = \frac{4}{2} = 2$
2. $x = \frac{3 - 1}{2} = \frac{2}{2} = 1$

So the solutions to the equation $x^2 - 3x + 2 = 0$ are $x = 2$ and $x = 1$.