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Question

If α,β are the roots of the quadratic equation x2+3x+6=0 , then find the equation whose roots are 1+α1α, 1+β1β


A

5y2+5y+2=0

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B

5y2+2y+5=0

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C

2y2+5y+5=0

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D

5y22y+5=0

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Solution

The correct option is A

5y2+5y+2=0


Let x is a root of the equation x2+3x+6=0 . We want to find the equation whose roots are 1+x1x
(Because the roots are 1+α1α, 1+β1β

Let it be y.

y1=1+x1x

y1y+1=1+x(1x)1+x+1x=x

Substitute y1y+1 in the equation

x2+3x+6 to get the equation whose root is y.

(y1y+1)2+3(y1y+1)+6=0(y1)2(y+1)2+3(y1)(y+1)(y+1)2+6(y+1)2(y+1)2=0y22y+1+3(y21)+6(y2+2y+1)=010y2+10y+4=0 or5y2+5y+2=0

(Replace y by x, because change of variable does not affect the equation)

We can solve this by finding the sum of the roots and product of the roots also.


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