3+4i is a root of x2+Ax+B=0 and 3−2 is a root of x2+Cx+D=0 then

If 3+4i is root of x2+Ax+B=0 Then 3−4i is also rootLet α=3+4i, β=3−4iα+β=−A⇒ A=−(3+4i+3−4i)=−6αβ= B⇒ B=(3+4i)(3−4i)=9+16= 25and If 3−2 root of x2+Cx+D=0 Then −2−3 is also rootr=3−2, δ=−2−3r+δ=−C⇒ C=−(3−2−2−3)=4rδ=D⇒ D= (3−2)(−2−3)=−(3−4)=1∴ A<D<C<B

3+4i is a root of x2+Ax+B=0 and 3−2 is a root of x2+Cx+D=0 then

If 3+4i is root of x2+Ax+B=0 Then 3−4i is also rootLet α=3+4i, β=3−4iα+β=−A⇒ A=−(3+4i+3−4i)=−6αβ= B⇒ B=(3+4i)(3−4i)=9+16= 25and If 3−2 root of x2+Cx+D=0 Then −2−3 is also rootr=3−2, δ=−2−3r+δ=−C⇒ C=−(3−2−2−3)=4rδ=D⇒ D= (3−2)(−2−3)=−(3−4)=1∴ A<D<C<B

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$3 + 4 i i s a r o o t o f x^{2} + A x + B = 0 a n d \sqrt{3} - 2 i s a r o o t o f x^{2} + C x + D = 0 t h e n$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$A > D > C > B$

$A > C > D > B$

$A < C < D < B$

$A < D < C < B$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} I f 3 + 4 i i s r o o t o f x^{2} + A x + B = 0 T h e n 3 - 4 i i s a l s o r o o t \\ L e t α = 3 + 4 i, β = 3 - 4 i \\ α + β = - A \Rightarrow A = - (3 + 4 i + 3 - 4 i) = - 6 \\ α β = B \Rightarrow B = (3 + 4 i) (3 - 4 i) = 9 + 16 = 25 \\ a n d I f \sqrt{3} - 2 r o o t o f x^{2} + C x + D = 0 T h e n - 2 - \sqrt{3} i s a l s o r o o t \\ r = \sqrt{3} - 2, δ = - 2 - \sqrt{3} \\ r + δ = - C \Rightarrow C = - (\sqrt{3} - 2 - 2 - \sqrt{3}) = 4 \\ r δ = D \Rightarrow D = (\sqrt{3} - 2) (- 2 - \sqrt{3}) = - (3 - 4) = 1 \\ ∴ A < D < C < B \end{array}$