If the system of equations r2+s2=t and r+s+t=k−32 has exactly one real solution, then the value of k is

Eliminating t, we get r+s+r2+s2=k−32 or  r2+r+s2+s+3−k2=0Now for one real solution of rD = 0⇒ 1−4s2+s+3−k2=0⇒ 4s2+4s+5−2k=0 Now for one real solution ' s ' D=0⇒ 16−16(5−2k)=0⇒ 1=5−2k⇒ k=2