If the roots of the equation ax2+bx+c=0 are of the form k+1k and k+2k+1 then (a+b+c)2 is equal to

We have,k+1k+k+2k+1=−ba                 (1)and k+1kk+2k+1=ca⇒ k+2k=ca or 2k=ca−1=c−aa or k=2ac−a               (2)Now, eliminate k. Putting the value of k in Eq. (1), we getc+a2a+2cc+a=−baor  (c+a)2+4ac=−2b(a+c)or  (a+c)2+2b(a+c)=−4acAdding b2 to both sides, we have(a+b+c)2=b2−4ac