If-the-equation-x-2-cx-d-0-has-roots-equal-to-the-fourth-powers-of-the-roots-of-x-2-ax-b-0-where-a-2-gt-4b-then-the-roots-of-x-2-4bx-2b-2-c-0-will-be-

Question Number 98058 by PengagumRahasiamu last updated on 11/Jun/20

If the equation x^{2} - c x + d = 0 has roots

equal to the fourth powers of the roots

of x^{2} + a x + b = 0, where a^{2} > 4 b then the

roots of x^{2} - 4 b x + 2 b^{2} - c = 0 will be

Answered by Rio Michael last updated on 11/Jun/20

Both real (one is positive and the other negative)

see how :

let x^{2} + a x + b = 0 have roots α and β

then the roots of x^{2} - c x + d = 0 are α^{4} and β^{4}

α + β = - a and α β = b

also α^{4} + β^{4} = c and α^{4} β^{4} = d

\Rightarrow b^{4} = d and α^{4} + β^{4} = c

{(α^{2} + β^{2})}^{2} - 2 {(α β)}^{2} = c

[{(α + β)}^{2} - 2 α β] - 2 {(α β)}^{2} = c

(a^{2} - 2 b^{2}) - 2 b^{2} = c \Rightarrow 2 b^{2} + c = {(a^{2} - 2 b)}^{2}

2 b^{2} - c = 4 a^{2} b - a^{2}

= a^{2} (4 b - a^{2})

now for x^{2} - 4 b x + 2 b^{2} - c = 0

discriminant D = {(4 b)}^{2} - 4 (1) (2 b^{2} - c)

= 16 b^{2} - 8 b^{2} + 4 c

= 8 b^{2} + 4 c

= 4 (2 b^{2} + c)

= 4 {(a^{2} - 2 b)}^{2} > 0 \Rightarrow real roots

f (0) = 2 b^{2} - c = a^{2} (4 b - a^{2}) < 0 \Rightarrow roots have opposite sign