Answer ths question plz.
Answers
Given Quadratic equation is abx^2 + acx + b(bx + c) = 0.
⇒ abx^2 + (ac + b^2)x + bc = 0.
On comparing with ax^2 + bx + c = 0, we get
a = ab, b = (ac + b^2), c = bc.
Given that the equation has real and equal roots.
⇒ Δ = b^2 - 4ac = 0
= (ac + b^2)^2 - 4(ab)(bc) = 0
= (a^2c^2 + b^4 + 2ab^2c) - 4ab^2c = 0
= a^2c^2 + b^4 + 2ab^2c - 4ab^2c = 0
= a^2c^2 + b^4 - 2ab^2c = 0
= (ac)^2 + (b^2)^2 - 2(ab)(b^2) = 0
= (ac - b^2)^2 = 0
= ac = b^2.
Option Verification:
(1) 4 & 49
⇒ b^2 = ac
= 4 * 49
= 196
b = 14.{Rational number}
(2) 49 & 16.
⇒ b^2 = ac
= 49 * 16
= 784
b = 28.{Rational number}
(3) 4 & 64
⇒ b^2 = ac
= 4 * 64
= 256
b = 16.{Rational number}
(4) 8 & 49.
⇒ b^2 = ac
= 392
b = 14√2, -14√2.{Irrational number}.
Therefore, the values of a and c respectively cannot be 8 & 49. - Option(D)
Hope it helps!