Question

12-1
If
V2+1
= +
x + y 2, the values of x and y are​

Answer 1

Answer:

Answer: The real values of x and y are

x = 5 and y = 2, -2.

Step-by-step explanation: We are given to find the real values of x and y or which the following equation holds :

(1+i)^2+(6+i)=(2+i)x~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)(1+i)

2

+(6+i)=(2+i)x (i)

We will be using the following value :

i^2=-1.i

2

=−1.

From equation (i), we have

\begin{gathered}(1+i)y^2+(6+i)=(2+i)x\\\\\Rightarrow y^2+iy^2+6+i=2x+ix\\\\\Rightarrow (y^2+6)+i(1+y^2)=2x+ix.\end{gathered}

(1+i)y

2

+(6+i)=(2+i)x

⇒y

2

+iy

2

+6+i=2x+ix

⇒(y

2

+6)+i(1+y

2

)=2x+ix.

Comparing the real and imaginary values from both sides of the above equation, we get

\begin{gathered}y^2+6=2x~~~\Rightarrow y^2=2x-6~~~~~~~~~~~~~~~~~~~~~~~~(ii)\\\\1+y^2=x~~~\Rightarrow y^2=x-1~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)\end{gathered}

y

2

+6=2x ⇒y

2

=2x−6 (ii)

1+y

2

=x ⇒y

2

=x−1 (iii)

Comparing equations (ii) and (iii), we get

\begin{gathered}2x-6=x-1\\\\\Rightarrow 2x-x=-1+6\\\\\Rightarrow x=5.\end{gathered}

2x−6=x−1

⇒2x−x=−1+6

⇒x=5.

From equation (iii), we get

\begin{gathered}y^2=5-1=4\\\\\Rightarrow y=\pm2.\end{gathered}

y

2

=5−1=4

⇒y=±2.

Thus, the real values of x and y are

x = 5 and y = 2, -2.