(a2+b2)x2-2b(a+c)x+(b2+c2)=0 For equal roots ...To find …
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Naveen Sharma 7 years, 2 months ago
Ans. On comparing with standard form.of quadratic equation
Ax2 + Bx + C = 0, We get
A = (a2 + b2)
B = -2b(a+c)
C = (b2 + c2)
as equation has equal roots,So
D = 0
=> B2 - 4AC = 0
=> [-2b(a+c)]2 - 4(a2 + b2)(b2 +c2) =0
=> 4b2(a2 + c2+2ac) = 4(a2b2 + a2c2 + b4 + b2c2)
divide by 4 both sides,
=> a2b2 + b2c2 - 2acb2 - a2b2 + a2c2 + b4 - b2c2 = 0
=> (ac)2 + (b2)2 - 2(ac)(b2) =0
=> (ac - b2)2 =0
squareroot both side
=> ac - b2 = 0
=> b2 = ac
Hence proved
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