SOLUTION: Let u and v be the solutions to 3x^2 + 5x + 7 = x^2 + 8x - 2. Find
1/u^3 + 1/v^3
Algebra.Com
Question 1209219: Let u and v be the solutions to 3x^2 + 5x + 7 = x^2 + 8x - 2. Find
1/u^3 + 1/v^3
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
Answer: -5/27
Explanation
turns into
u^3 + v^3 = (u + v)(u^2 - uv + v^2) is the sum of cubes factoring rule
Use the identity u^2+v^2 = (u+v)^2-2uv to rewrite that previous equation like so
u^3 + v^3 = (u + v)(u^2 - uv + v^2)
u^3 + v^3 = (u + v)(u^2 + v^2 - uv)
u^3 + v^3 = (u + v)((u+v)^2-2uv - uv)
u^3 + v^3 = (u + v)((u+v)^2 - 3uv)
At this point we have terms involving u+v and uv
What can we do with this? We can use Vieta's Formulas.
But first 3x^2 + 5x + 7 = x^2 + 8x - 2 must be rearranged into 2x^2-3x+9 = 0
Then divide everything by the leading coefficient to get x^2-(3/2)x+9/2 = 0
Due to Vieta's formulas, the roots add to the negative of the x coefficient and multiply to the constant term. This applies only when the leading coefficient is 1.
We have these equations
u+v = 3/2
u*v = 9/2
So,
u^3 + v^3 = (u + v)((u+v)^2 - 3uv)
u^3 + v^3 = (3/2)*( (3/2)^2 - 3*(9/2) )
u^3 + v^3 = -135/8
And,
=
=
=
Therefore,
where u,v are the roots of 3x^2 + 5x + 7 = x^2 + 8x - 2
I used GeoGebra to verify the answer is correct.
-5/27 = -0.185185 approximately where the "185" repeats forever.
More practice is found here
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