For a, b, c ∈ R- 0 let (a+b/1-a b), b, (b+c/1-b c) are in A.P. If α, β are the roots of the quadratic equation 2 a c x2+2 a b c x+(a+c)=0, then find the value of (1+α)(1+β).

Question

For a, b, c ∈ R- 0 let (a+b/1-a b), b, (b+c/1-b c) are in A.P. If α, β are the roots of the quadratic equation 2 a c x2+2 a b c x+(a+c)=0, then find the value of (1+α)(1+β). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Given (a+b/1-a b), b, (b+c/1-b c) are in A.P. ⇒ b -( a + b /1- ab )=( b + c /1- bc )- b ⇒ (- a ( b 2+1)/1- ab )=( c ( b 2+1)/1- bc ) ⇒ a + c =2 abc Now, given quadratic equation is 2 a c x2+2 a b c x+2 a b c=0 (Substituting a+c=2 a b c and then cancelling 2ac) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/ea6f9264e8559f80cdc32a16f475e4c3-.png" /> As α+β=-b, α β=b ∴ (1+α)(1+β)=(α+β)+(α β)+1=-b+b+1=1

Q. For a,b,c∈R−{0}, let 1−aba+b​,b,1−bcb+c​ are in A.P. If α,β are the roots of the quadratic equation 2acx2+2abcx+(a+c)=0, then find the value of (1+α)(1+β).

Given 1−aba+b​,b,1−bcb+c​ are in A.P. ⇒b−1−aba+b​=1−bcb+c​−b⇒1−ab−a(b2+1)​=1−bcc(b2+1)​⇒a+c=2abc Now, given quadratic equation is 2acx2+2abcx+2abc=0 (Substituting a+c=2abc and then cancelling 2ac) As α+β=−b,αβ=b ∴(1+α)(1+β)=(α+β)+(αβ)+1=−b+b+1=1

Q. For $a, b, c \in R - {0}$ , let $\frac{a + b}{1 - ab}, b, \frac{b + c}{1 - b c}$ are in A.P. If $α, β$ are the roots of the quadratic equation $2 a c x^{2} + 2 ab c x + (a + c) = 0$ , then find the value of $(1 + α) (1 + β)$ .