For a,b,c non-zero, real distinct, the equation, a2+b2x2−2b(a+c)x+b2+c2=0 has non-zero real roots.One of these roots is also the root of the equation:
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a
b2−c2x2+2a(b−c)x−a2=0
b
b2+c2x2−2a(b+c)x+a2=0
c
a2x2+a(c−b)x−bc=0
d
a2x2−a(b−c)x+bc=0
answer is C.
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Detailed Solution
a2+b2x2−2b(a+c)x+b2+c2=0D=4b2(a+c)2−4a2+b2b2+c2 =−4b4−2b2ac+a2c2=−4b2−ac2 For real roots, D≥0⇒−4b2−ac2≥0⇒ b2−ac=0⇒ Roots are real and equal. ∴ Roots are 2b(a+c)±02a2+b2=b(a+c)a2+ac=baThis root satisfies option (c).