Question

simple question

solve it

thanks you !!

Answer 1

1 The expression {y}^{2}-2y+2y
​2
​​ −2y+2 fits the form a{x}^{2}+bx+cax
​2
​​ +bx+c, where:
a=1a=1
b=-2b=−2
c=2c=2

2 Introduce the constant kk, which is 11 in our case
1 Let kk = {(\frac{b}{2a})}^{2}(
​2a
​
​b
​​ )
​2
​​
From the earlier step, a=1a=1 and b=-2b=−2
Therefore, k={(\frac{b}{2a})}^{2}={(-\frac{2}{2\times 1})}^{2}=1k=(
​2a
​
​b
​​ )
​2
​​ =(−
​2×1
​
​2
​​ )
​2
​​ =1
k=1k=1
{y}^{2}-2y+1-1+2y
​2
​​ −2y+1−1+2
3 Use Square of Difference: {(a-b)}^{2}={a}^{2}-2ab+{b}^{2}(a−b)
​2
​​ =a
​2
​​ −2ab+b
​2
​​
{(y-1)}^{2}-1+2(y−1)
​2
​​ −1+2
4 Simplify
{(y-1)}^{2}+1(y−1)
​2
​​ +1
5 Substitute the above back into the original equation
{(y-1)}^{2}+1=0(y−1)
​2
​​ +1=0
6 Subtract 11 from both sides
{(y-1)}^{2}=-1(y−1)
​2
​​ =−1
7 Take the square root of both sides
y-1=\pm \sqrt{-1}y−1=±√
​−1
​

8 Simplify \sqrt{-1}√
​−1

​​ to \sqrt{1}\imath √
​1
​
​​ ı
y-1=\pm \sqrt{1}\imath y−1=±√
​1
​
​​ ı

9 Simplify \sqrt{1}√
​1
​
​​ to 11
y-1=\pm 1\times \imath y−1=±1×ı

10 Simplify 1\times \imath 1×ı to \imath ı
y-1=\pm \imath y−1=±ı

11 Add 11 to both sides
y=\imath +1,-\imath +1y=ı+1,−ı+1

12 Regroup terms
y=1+\imath ,-\imath +1y=1+ı,−ı+1

13 Regroup terms
y=1+\imath ,1-\imath y=1+ı,1−

Answer 2

here is your answer hope it will help you