Question

1/7x^2-2x+7=0Квпд ур​

Answer 1

Ответ: x=7

Объяснение:

$\frac{1}{7}x^2-2x+7=0\ \ \ |\cdot7\\\\x^2-14x+49=0\\\\(x-7)^2=0\\\\x-7=0\\\\x=7$

Answer 2

To solve the quadratic equation $\frac{1}{7}x^2 - 2x + 7 = 0$, 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 our equation, $a = \frac{1}{7}$, $b = -2$, and $c = 7$.

Substitute the values into the formula:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot \frac{1}{7} \cdot 7}}{2 \cdot \frac{1}{7}}$

Simplifying further:

$x = \frac{2 \pm \sqrt{4 - 4 \cdot \frac{1}{7} \cdot 7}}{\frac{2}{7}}$ $x = \frac{2 \pm \sqrt{4 - 4}}{\frac{2}{7}}$ $x = \frac{2 \pm \sqrt{0}}{\frac{2}{7}}$

Since the discriminant (the term inside the square root) is zero, we have two identical real solutions:

$x = \frac{2}{\frac{2}{7}} = 7$

So, the solution to the quadratic equation is $x = 7$.