Asked by Patrick
If α and β are the root of the equation 2X²-7X-3=0 Find
α
---- + 1 +
β
β
---- + 1
α
α
---- + 1 +
β
β
---- + 1
α
Answers
Answered by
Reiny
In this forum we type fractions as x/y
so did you mean:
α/β + 1 + β/α + 1
or
(α + 1)//β + (β + 1)/α
While you decide, remember that
α + β = 7/2
αβ = -3/2
so did you mean:
α/β + 1 + β/α + 1
or
(α + 1)//β + (β + 1)/α
While you decide, remember that
α + β = 7/2
αβ = -3/2
Answered by
Just saw your previous .... Reiny
So you wanted:
α/(β + 1) + β/(α + 1 )
As you can see skipping those brackets causes major problems
in interpretation. I had not even considered this interpretation.
Using the sum and product property of a quadratic makes the
solution less complicated
α/(β + 1) + β/(α + 1 ) , I will switch to a and b instead of the greek letters.
= (a(a+1) + b(b+1))/(ab +a + b + 1))
= (a^2 + a + b^2 + b)/(ab+a+b+1)
= ( (a+b)^2 - 2ab + a+b)/(ab + a+b + 1)
= ( 49/4 - 2(-3/2) + 7/2)/(-3/2 + 7/2 + 1)
= (75/4) / 3
= 25/4 = 6.25
Damon got 150/48 or 25/8
but if you take his answers to the equation of (7 ± √73)/4
you do get 24/4
α/(β + 1) + β/(α + 1 )
As you can see skipping those brackets causes major problems
in interpretation. I had not even considered this interpretation.
Using the sum and product property of a quadratic makes the
solution less complicated
α/(β + 1) + β/(α + 1 ) , I will switch to a and b instead of the greek letters.
= (a(a+1) + b(b+1))/(ab +a + b + 1))
= (a^2 + a + b^2 + b)/(ab+a+b+1)
= ( (a+b)^2 - 2ab + a+b)/(ab + a+b + 1)
= ( 49/4 - 2(-3/2) + 7/2)/(-3/2 + 7/2 + 1)
= (75/4) / 3
= 25/4 = 6.25
Damon got 150/48 or 25/8
but if you take his answers to the equation of (7 ± √73)/4
you do get 24/4
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