BITSAT 2022
Exam BITSAT
Year 2022
Subject Mathematics
Chapter Quadratic Equation BITSAT 2022 · Mathematics · Quadratic Equation
Let a, b be the solutions of x^2+p x+1=0 and c, d be the solution...
Let
a , b a, b a , b be the solutions of
x 2 + p x + 1 = 0 x^2+p x+1=0 x 2 + p x + 1 = 0 and
c , d c, d c , d be the solution of
x 2 + q x + 1 = 0 x^2+q x+1=0 x 2 + q x + 1 = 0 . If
( a − c ) ( b − c ) (a-c)(b-c) ( a − c ) ( b − c ) and
( a + d ) ( b + d ) (a+d)(b+d) ( a + d ) ( b + d ) are the solution of
x 2 + a x + β = 0 x^2+a x+\beta=0 x 2 + a x + β = 0 , then
β \beta β is equal to
Options A p + q p+q p + q B p − q p-q p − q C p 2 + q 2 p^2+q^2 p 2 + q 2 D q 2 − p 2 q^2-p^2 q 2 − p 2 Correct answer D. q 2 − p 2 q^2-p^2 q 2 − p 2
Step-by-step solution
Since,
a + b = − p , a b = 1 a+b=-p, a b=1 a + b = − p , ab = 1 .....(i)
a n d c + d = − q , c d = 1 and c+d=-q, c d=1 an d c + d = − q , c d = 1 Now
( a − c ) ( b − c ) a n d ( a + d ) ( b + d ) (a-c)(b-c) and (a+d)(b+d) ( a − c ) ( b − c ) an d ( a + d ) ( b + d ) are the roots of
x 2 + a x + β = 0 x^2+a x+\beta=0 x 2 + a x + β = 0 ( a − c ) ( b − c ) ( a + d ) ( b + d ) = β ⇒ ( a b − a c − b c + c 2 ) ( a b + a d + b d + d 2 ) = β ⇒ { 1 − c ( a + b ) + c 2 } { 1 + d ( a + b ) + d 2 } = β ⇒ ( 1 + p c + c 2 ) ( 1 − p d + d 2 ) = β ⇒ 1 − p d + d 2 + p c − p 2 c d + p c d 2 + c 2 − p c 2 d + c 2 d 2 = β ⇒ 1 − p d + d 2 + p c − p 2 + p d + c 2 − p c + 1 = β [ ∵ c d = 1 ] ⇒ 2 + d 2 + c 2 − p 2 = β \begin{aligned}& (a-c)(b-c)(a+d)(b+d)=\beta\\ & \Rightarrow\left(a b-a c-b c+c²\right)\left(a b+a d+b d+d²\right)=\beta\\ & \Rightarrow\left\{1-c(a+b)+c²\right\}\left\{1+d(a+b)+d²\right\}=\beta\\ & \Rightarrow \quad\left(1+p c+c²\right)\left(1-p d+d²\right)=\beta\\ & \Rightarrow 1-p d+d²+p c-p² c d+p c d²+c²-p c² d \\ & +c² d²=\beta\\ & \Rightarrow 1-p d+d²+p c-p²+p d+c²-p c+1=\beta\\ & {[\because c d=1]} \\ & \Rightarrow \quad 2+d²+c²-p²=\beta\\ & \end{aligned} ( a − c ) ( b − c ) ( a + d ) ( b + d ) = β ⇒ ( ab − a c − b c + c 2 ) ( ab + a d + b d + d 2 ) = β ⇒ { 1 − c ( a + b ) + c 2 } { 1 + d ( a + b ) + d 2 } = β ⇒ ( 1 + p c + c 2 ) ( 1 − p d + d 2 ) = β ⇒ 1 − p d + d 2 + p c − p 2 c d + p c d 2 + c 2 − p c 2 d + c 2 d 2 = β ⇒ 1 − p d + d 2 + p c − p 2 + p d + c 2 − p c + 1 = β [ ∵ c d = 1 ] ⇒ 2 + d 2 + c 2 − p 2 = β ⇒ 2 c d + c 2 + d 2 − p 2 = β ( c + d ) 2 − p 2 = β q 2 − p 2 = β [ ∵ ( c + d ) = − q ] \begin{aligned} \Rightarrow \quad 2 c d+c^2+d^2-p^2 & =\beta \\ (c+d)^2-p^2 & =\beta \\ q^2-p^2 & =\beta \quad[\because(c+d)=-q]\end{aligned} ⇒ 2 c d + c 2 + d 2 − p 2 ( c + d ) 2 − p 2 q 2 − p 2 = β = β = β [ ∵ ( c + d ) = − q ] $$
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