Question

Найдите корни уравнения (х-2)^2-1=0

Answer 1

(х-2)²-1=0
х²-4х+4-1=0
х²-4х+3=0
Дискриминант: B²-4AC ⇒ 16-12=4
$x= \frac{4-2}{2} =1$
$x= \frac{4+2}{2} = 3$
Ответ: х=1; 3

Answer 2

( Х - 2 ) ^ 2 - 1 = 0
Х^2 - 4х + 4 - 1 = 0
Х^2 - 4х + 3 = 0
D = 16 - 12 = 4 ; V D = 2
X1 = ( 4 + 2 ) : 2 = 3
X2 = ( 4 - 2 ) : 2 = 1
Ответ 3 и 1

Answer 3

I can help you with math problems and other topics. I see that you are asking me to find the roots of the equation $(x-2)^2-1=0$ in detail. Here is how I would solve it:

First, I would expand the left side of the equation using the formula $(a-b)^2=a^2-2ab+b^2$. In this case, $a=x$ and $b=2$, so I get:

$(x-2)^2-1=x^2-4x+4-1=x^2-4x+3$

Then, I would use the quadratic formula to find the roots of the equation $ax^2+bx+c=0$. The formula is:

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

In this case, $a=1$, $b=-4$, and $c=3$, so I get:

$x=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(3)}}{2(1)}=\frac{4\pm\sqrt{16-12}}{2}=\frac{4\pm\sqrt{4}}{2}=\frac{4\pm2}{2}$

Therefore, the roots are:

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

You can check these answers by plugging them back into the original equation and verifying that they make it true.

I hope this helps you understand how to solve quadratic equations. If you want to learn more, you can visit some of these websites that explain the topic in more detail. Have a nice day!