If the equation a(b–c)x2+b(c–a)x+c(a–b)=0 has equal roots, where a + c = 15 and b=365, then a2+c2 is equal to __________. [2025]
(117)
Since, a(b–c)x2+b(c–a)x+c(a–b)=0 has equal roots
Put x = 1 in given equation.
∴ ab – ac + bc – ab + ac – bc = 0
∴ Both roots of the equation is 1.
∴ 1+1=–b(c–a)a(b–c)
⇒ 2ab–2ac=–bc+ab
⇒ 2ac=bc+ab ⇒ 2ac=b(a+c)
⇒ 2ac=365×15=108
Also, a + c = 15 [Given]
∴ a2+c2+2ac=225
⇒ a2+c2=225–108 ⇒ a2+c2= 117.