AFCAT Model Paper 1

Given:
x=√3+√2
Formula used:
(x+

1

x

)(x−

1

x

)=x2−

1

x2

Calculation:
x=√3+√2 ...(i)
⇒

1

x

=

1

(√3+√2)

⇒ [

1

(√3+√2)

]×[

(√3−√2)

(√3−√2)

]
⇒ (√3−√2)
⇒

1

x

=(√3−√2) ...(ii)
On adding (i) and (ii), we get
[x+(

1

x

)]=√3+√2+√3−√2
⇒ [x+(

1

x

)]=2√3 ...(iii)
On subtracting (ii) by (i), we get
x–(

1

x

)=√3+√2−√3+√2
⇒ x–(

1

x

)=2√2
Now, we have to find the value of x2–(

1

x2

)
x2−(

1

x2

)=[x−(

1

x

)][x+(

1

x

)]
From equation (ii) & (iii)
⇒ 2√2×2√3
⇒ 4√6
∴ The value of x2−(

1

x2

) is 4√6.