(a2+b2)x2-2b(a+c)x+(b2+c2)=0For equal roots ...To find b2=????

1.	(a2+b2)x2-2b(a+c)x+(b2+c2)=0For equal roots ...To find b2=????
Answer» Ans. On comparing with standard form.of quadratic equation\xa0Ax2\xa0+ Bx + C = 0, We getA = (a2\xa0+ b2)B = -2b(a+c)C = (b2\xa0+ c2)as equation has equal roots,SoD = 0=> B2\xa0- 4AC = 0=> [-2b(a+c)]2\xa0- 4(a2\xa0+ b2)(b2\xa0+c2) =0=> 4b2(a2\xa0+ c2+2ac) = 4(a2b2\xa0+ a2c2\xa0+ b4\xa0+ b2c2)divide by 4 both sides,=> a2b2\xa0+ b2c2\xa0- 2acb2\xa0- a2b2\xa0+ a2c2\xa0+ b4\xa0- b2c2\xa0= 0=> (ac)2\xa0+ (b2)2\xa0- 2(ac)(b2) =0=> (ac - b2)2\xa0=0squareroot both side=> ac - b2\xa0= 0=> b2\xa0= acHence proved

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