If α and β are the roots of the equation 2x^2 + 2px + p^2 = 0 where, p is a non-zero real number and \alpha^4 and \beta^4 are the roots of x^2 - rx + s = 0, then the roots of 2x^2 - 4p^2 x + 4p - 2r = 0 are

Correct Answer is : equal and non-zero

NIMCET 2014 Question Paper with solutions

Question : 16 of 120

Marks: +1, -0

If α and β are the roots of the equation

2x^2

+ 2px +

p^2

= 0 where, p is a non-zero real number and

α^4

and

β^4

are the roots of

x^2

- rx + s = 0, then the roots of

2x^2 - 4p^2 x

+ 4p - 2r = 0 are

Solution:

We have,

2x^2 + 2px + p^2

= 0

α + β = - p

αβ = p^2/2 ... (i)

and

α^4

and

β^4

are the roots

x^2 - rx + 3 = 0

α^4+β^4 = r

α^4β^4 = s ... (ii)

Now ,

2x^2 - 4p^2x + 4p - 2r = 0

D = B^2 - 4AC

∴D = 16p^4 - 4 [2(p^2-2r)]

= 16p^4 - 8p^4+16r

= 8 (p^4+2r)

= 8 [4α^2β^2+2(α^4+β^4)]

= 16 [(α+β)^2-2αβ]^2

= 16 (p^2-p^2) = 0

Hence, the roots of the equation

2x^2-4p^2 x + 4p - 2r = 0

are equal

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