Roots of an Interesting Polynomial
Suppose the roots of x 3 + 3 x 2 + 4 x − 1 1 are α , β , γ .
If the roots of x 3 + r x 2 + s x + t are α + β , α + γ , β + γ , then find t .
The answer is 23.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
This is actually the best as you can also get the value of r and s. You should have received the most upvotes.
This is the way for solving such problems.
By Vieta's formulas, we have:
⎩ ⎪ ⎨ ⎪ ⎧ α + β + γ = − 3 α β + β γ + γ α = 4 α β γ = 1 1
And, we have:
t = − ( α + β ) ( β + γ ) ( γ + α ) = − ( − 3 − γ ) ( − 3 − β ) ( − 3 − α ) = ( 3 + α ) ( 3 + β ) ( 3 + γ ) = 2 7 + 9 ( α + β + γ ) + 3 ( α β + β γ + γ α ) + α β γ = 2 7 + 9 ( − 3 ) + 3 ( 4 ) + 1 1 = 2 3
we are looking for t = − ( α + β ) ( β + γ ) ( γ + α ) by daniel lui identity: t = − ( ( α + β + γ ) ( α β + β γ + γ α ) − α β γ ) substitute from vietas t = − ( − 4 ∗ 3 − 1 1 ) = 2 3
For the sake of variety, I'm gonna use transformation of roots. α + β = − 3 − γ . So we are looking for a polynomial with its roots as − 3 − α , − 3 − β , − 3 − γ . This polynomial can be obtained by replacing the x in x 3 + 3 x 2 + 4 x − 1 1 with − 3 − x . On solving, we get t = 2 3 .