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Question

If (α+β) and (α-β) are the roots of the equation x2+px+q=0, where α,β,p and q are real, then the roots of the equation (p2-4q)(p2x2+4px)-16q=0 are


A

1ɑ+1β and 1ɑ-1β

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B

1ɑ+1β and 1ɑ-1β

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C

1ɑ+1β and 1ɑ-1β

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D

ɑ+β and ɑ-β

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Solution

The correct option is A

1ɑ+1β and 1ɑ-1β


Explanation for the correct option:

Step 1. Find the roots of (p2-4q)(p2x2+4px)-16q=0:

Given, (α+β) and (α-β) are the roots of the equation x2+px+q=0

Sum of the roots =-p

(α+β)+(αβ)=p

2α=p

α=p2

and product of roots =q

(α+β)(αβ)=q

α2β=q

Step 2. Put the value of α2 in α2β=q

p24β=q

β=p24q4

Step 3. put the values of β and α2 in (p2-4q)(p2x2+4px)-16q=0, we get

(p2-4q)(p2x2+4px)-16q=0

4β(4α2x28αx)16(α2β)=0

β(4α2x28αx)4(α2β)=0

α2βx22αβx+β=α

αβxβ2=α2

αβxβ=±α

x=1α±1β

The required roots are x=1ɑ+1β and 1ɑ-1β

Hence, Option ‘A’ is Correct.


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