" 18."a^(2)-b^(2)-4ac+4c^(2)

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

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)

if alpha , beta are root of ax^2+bx+c=0 then (1/alpha^2+1/beta^2)^2 (a) (b^(2)(b^(2)-4ac))/(c^(2)a^(2)) (b) (b^(2)(b^(2)-4ac))/(ca^(3)) (c) (b^(2)(b^(2)-4ac))/(a^(4)) (d) (b^(2)-2ac)^2/(c^(4))

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

If the roots of the equation ax^(2)+bx+c=0 are of the form (k+1)/k and (k+2)/(k+1), then (a+b+c)^(2) is equal to 2b^(2)-ac b.a62 c.b^(2)-4ac d.b^(2)-2ac

If a:b=b:c,(:thenbackslash a^(4):b^(4) would be equal to ac:b^(2) b.a^(2):c^(2) c.c^(2):a^(2) d.b^(2):ac

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

The sides of DeltaABC 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)