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The roots of the equation \[2x^2 + 5x - 8 = 0\] are \[\alpha \] and \[\beta \]. Find the quadratic equations whose roots are : \[5\alpha + \frac{ 1 }{ \alpha }, 5\beta + \frac{ 1 }{ \beta } \]
Have you found the sum and product of them?

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Other answers:

no
I did this :
\[\alpha+\beta = \frac{ -5 }{ 2 } \] , \[\alpha \beta = \frac{ -8 }{ 2 } = -4\] New sum = \[5\alpha + \frac{ 1 }{ \alpha } + 5\beta + \frac{ 1 }{ \beta } \] I don't know what to do now
well why not make it \[\frac{5\alpha^2 + 1}{\alpha} + \frac{5\beta^2 + 1}{\beta}\]
Then you take the sum and product of that?
then get the common denominator and add the numerators
then for new product: \[(5\alpha + \frac{ 1 }{ \alpha })* (5\beta+\frac{ 1 }{ \beta })\] in the end I got the answer to be : \[8x^2 -95 - 786\]. which I believe is wrong :( !!!
then input the sum and product into \(\sf x^2 -(\text{sum of roots})x +(\text{product of roots})=0\)?
look at the sum of the roots this way \[5\alpha + 5 \beta = \frac{1}{\alpha} + \frac{1}{\beta} = 5(\alpha + \beta) + \frac{\alpha + \beta}{\alpha \beta}\] does that help
oops should read \[5 \alpha + 5 \beta + \frac{1}{\alpha} + \frac{1}{\beta}\]
ahh yes that does help very much .. cheers :)
so was I correct then ????
what did you guys get as your final answer???
5(-5/2) + (-5/2)/-4 = -100/8 + 5/8 = -95/8 so that seems correct
what about the product ... it seems way tooo big
as for the product I thought \[(5\alpha + \frac{1}{\alpha})(5\beta + \frac{1}{\beta} = 25\alpha \beta + \frac{5\alpha}{\beta} + \frac{5\beta}{\alpha} + \frac{1}{\alpha \beta}\]
which becomes \[25 \alpha \beta + \frac{5 \alpha^2 + 5\beta^2}{\alpha \beta} + \frac{1}{\alpha \beta}\] or \[25 \alpha \beta + \frac{5[(\alpha + \beta)^2 - 2 \alpha \beta]}{\alpha \beta} + \frac{1}{\alpha \beta}\]
so I think you need to be careful with the signs
tooo confusing wht do you get as finl answer ???
i got -4
ok ....
so if -b/a = -95/8 then c/a = -32/8 so a = 8, b = 95 and c -32 that's my best guess
I got something else

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