If the system of equation `r^(2)+s^(2)=t` and `r+s+t=(k-3)/(2)` has exactly one real solution, then the value of `k` is

Options

A. `1`
B. `2`
C. `3`
D. `4`

Answer

B

Explanation

`(b)` Eliminating `t`, we get
`r+s+r^(2)+s^(2)=(k-3)/(2)`
or `r^(2)+r+s^(2)+s+(3-k)/(2)=0`
Now for one real solution `'r'`
`D=0`
`:. 1-4(s^(2)+s+(3-k)/(2))=0`
`:. 4s^(2)+4s+5-2k=0`
Now for one real solution `'s'`
`D=0`
`:. 16-16(5-2k)=0`
`:. 1=5-2k`
`:. k=2`