If the roots of ax2+bx+c=0  are α,  β and the roots of Ax2+Bx+C=0  are α−k,  β−k  then (B2−4AC)(b2−4ac) is equal to

The given equation is ax2+bx+c=0  …… ( 1 )α,  β and the roots of Eq. ( 1 )Sum of the roots α+β=−baProduct of the roots αβ=caNow (α−β)2=(α+β)2−4αβ(α−β)2 =(b2−4ac)a2        …… ( 2 )Another given equation is Ax2+Bx+C=0      …… ( 3 )α−k,  β−k  are the roots of Eq. ( 3 )Also[(α−k)−(β−k)]2 ={(α−k)+(β−k)}2−4(α−k)(β−k)=(−BA)2−4(CA)=(B2−4AC)A2        ……(4)From (2) and (4),(b2−4ac)a2=(B2−4AC)A2∴ B2−4ACb2−4ac=(Aa)2