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Algebra Level 5

Let r , s r,s and t t be the roots of x 3 + x A + x 2 A + A = 0 x^3+ x\sqrt{A} + x\sqrt{2A} + A =0 . Find ( r + s + t ) ( r + s − t ) r 2 + s 2 − t 2 + 2 r s \frac{(r+s+t)(r+s-t)}{r^2+s^2-t^2+2rs} where 100 ≤ A ≤ − 100 100 \leq A \leq -100 .

Image Credit: Wikimedia Francois Viete
None of the choices 5 + 1 \sqrt{5}+1 0 0 1 1

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Paul Ryan Longhas by Paul Ryan Longhas , 24, Philippines

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2 solutions

Paul Ryan Longhas
Feb 27, 2015

x 3 + x A + x 2 A + A = 0 = > r + s + t = 0 x^3+ x\sqrt{A} + x\sqrt{2A} + A =0 => r+s+t =0 (By Vieta)

Also, ( r + s + t ) ( r + s − t ) = r 2 + s 2 − t 2 + 2 r s = 0 (r+s+t)(r+s-t) = r^2+s^2-t^2+2rs = 0

Hence, ( r + s + t ) ( r + s − t ) r 2 + s 2 − t 2 + 2 r s = ( r + s + t ) ( r + s − t ) ( r + s + t ) ( r + s − t ) \frac{(r+s+t)(r+s-t)}{r^2+s^2-t^2+2rs} = \frac{(r+s+t)(r+s-t) }{(r+s+t)(r+s-t)}

So, the expression will become 0 0 \frac{0}{0} wich is not defined.

Therefore, the answer is none of the choices .

7 Helpful 0 Interesting 0 Brilliant 0 Confused

@Paul Ryan Longhas Don't u think that q is overrated

Mehul Chaturvedi - 6 years, 3 months ago
Archit Boobna
Mar 25, 2015

It is impossible to have a number A where 100 ≤ A ≤ − 100 100\le A\le -100 . There was no need to solve the question

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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