Question

125x²-5=0 Знайти корені рівняння

Answer 1

125x²-5=0
5(25x²-1)=0
(5x-1)(5x+1)=0
5x-1=0   x=1/5
5x+1=0   x=-1/5

ответ {-1/5;1/5}

Answer 2

125x²-5=0
5(25х²-1)=0
(5х-1)(5х+1)=0
5х-1=0
5х=1
х=1:5
х=0,2

5х+1=0
5х=-1
х=-1:5
х=-0,2
Ответ:-0,2; 0,2

Answer 3

To find the roots of the quadratic equation $125x^2 - 5 = 0$, we can use the quadratic formula, which is given by:

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

where the equation is in the form $ax^2 + bx + c = 0$.

In this case, $a = 125$, $b = 0$, and $c = -5$.

Let's plug these values into the quadratic formula to find the roots:

$x = \frac{-(0) \pm \sqrt{(0)^2 - 4 \cdot 125 \cdot (-5)}}{2 \cdot 125}$

$x = \frac{\pm \sqrt{0 + 2500}}{250}$

$x = \frac{\pm \sqrt{2500}}{250}$

$x = \frac{\pm 50}{250}$

Now, we have two possible solutions for $x$:

1. $x = \frac{50}{250} = \frac{1}{5}$
2. $x = \frac{-50}{250} = -\frac{1}{5}$

So, the roots of the quadratic equation $125x^2 - 5 = 0$ are $x = \frac{1}{5}$ and $x = -\frac{1}{5}$.