Q.

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]


Ans.

(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.