a^(2)-b^(2)-4ac+4c^(2)

Factorise : a^(2)-9b^(2)+4c^(2)-25d^(2)-4ac+30bd

Let alpha and beta be the roots of the equationa x^(2)+2bx+c=0 and alpha+gamma and beta+gamma be the roots of Ax^(2)+2Bx+C=0. Then prove that A^(2)(b^(2)-4ac)=a^(2)(B^(2)-4AC)

The sides of DeltaABC satisfy the equation 2a^(2) + 4b^(2) + c^(2) = 4ab + 2ac . Then

The sides of a triangleABC satisfy the equation 2a^(2) + 4b^(2) + c^(2) =4ab + 2ac , then-

Prove that {:[( a^(2) , bc, ac+c^(2)),( a^(2) +ab,b^(2) ,ac),( ab,b^(2) +bc,c^(2)) ]:} =4a^(2) b^(2) c^(2)

Prove that {:|( a^(2) , bc, ac+c^(2)),( a^(2) +ab,b^(2) ,ac),( ab,b^(2) +bc,c^(2)) |:} =4a^(2) b^(2) c^(2)

Prove that {:|( a^(2) , bc, ac+c^(2)),( a^(2) +ab,b^(2) ,ac),( ab,b^(2) +bc,c^(2)) |:} =4a^(2) b^(2) c^(2)

Prove that {:[( a^(2) , bc, ac+c^(2)),( a^(2) +ab,b^(2) ,ac),( ab,b^(2) +bc,c^(2)) ]:} =4a^(2) b^(2) c^(2)

Using properties of determinants, prove the following abs{:(a^2, bc, ac +c^2 ),(a^(2) + ab, b^(2),ac ),(ab, b^(2) + bc,c^(2) ):}=4a^(2) b^(2) c^(2) .

Four steps to derive the quadratic formula are shown below . (I) x^(2)+(bx)/a=(-c)/a (II) (x+b/(2a))^(2)=(b^(2)-4ac)/(4a^(2)) (III) x =pm sqrt((b^(2)-4ac)/(4a^(2)))-b/(2a) (IV) x^(2)+(bx)/a +(b/(2a))^(2)=(-c)/a + (b/(2a))^(2) What is the correct order for these steps ?