Asked by Patrick
If α and β are the root of the equation 2X²-7X-3=0 Find α/β+1 + β/α+1
Working answer please
Working answer please
Answers
Answered by
Anonymous
x = [ 7 +/- sqrt (49 + 24) ] / 4
α = (7+sqrt 73)/4
β = (7-sqrt 73)/4
now I do not care to go on because I am not sure if you mean what you typed
(α/β)+1
or
α/ (β+1)
α = (7+sqrt 73)/4
β = (7-sqrt 73)/4
now I do not care to go on because I am not sure if you mean what you typed
(α/β)+1
or
α/ (β+1)
Answered by
Patrick
It is α/β+1 + β/α+1
That is how the question is
That is how the question is
Answered by
Damon
But it is not typed on one line in your book.
What is the numerator and what is the denominator?
You need more parentheses when typing online.
What is the numerator and what is the denominator?
You need more parentheses when typing online.
Answered by
Damon
a
---- + 1
b
or
a
------
b+1
---- + 1
b
or
a
------
b+1
Answered by
Patrick
α/β + 1 + β/α + 1
Sorry its error in tying
Sorry its error in tying
Answered by
Patrick
α
---- + 1
β
---- + 1
β
Answered by
Damon
α/β + 1 + β/α + 1 = α/β + β/α + 2
I still think you may have it wrong
are you sure you do not mean:
α/(β + 1)
+
β/(α + 1 )
I still think you may have it wrong
are you sure you do not mean:
α/(β + 1)
+
β/(α + 1 )
Answered by
Patrick
Okay .. Please you can still help out with
α/(β + 1)
+
β/(α + 1 )
α/(β + 1)
+
β/(α + 1 )
Answered by
Damon
β+1 = (7-sqrt 73)/4 + 4/4 = (11-sqrt 73)/4
α+1 = (7 +sqrt 73)/4 + 4/4 = (11+sqrt 73)/4
α/(β + 1) = (7+sqrt 73)/(11-sqrt73)
β/(α + 1 ) = (7-sqrt 73)/(11+sqrt 73)
α/(β + 1) = (7+sqrt 73)/(11-sqrt73) * (11+sqrt73)/(11+sqrt73)
β/(α + 1 ) = (7-sqrt 73)/(11+sqrt 73) * (11-sqrt 73)/(11-sqrt 73)
α/(β + 1) = (7+sqrt 73)* (11+sqrt73)/(121-73)
β/(α + 1 ) = (7-sqrt 73) * (11-sqrt 73)/(121- 73)
(77 +18 sqrt 73 +73)/48
(77 - 18 sqrt 73 +73/48
-------------------------------- add
(154 + 146) /48
= 150/48
α+1 = (7 +sqrt 73)/4 + 4/4 = (11+sqrt 73)/4
α/(β + 1) = (7+sqrt 73)/(11-sqrt73)
β/(α + 1 ) = (7-sqrt 73)/(11+sqrt 73)
α/(β + 1) = (7+sqrt 73)/(11-sqrt73) * (11+sqrt73)/(11+sqrt73)
β/(α + 1 ) = (7-sqrt 73)/(11+sqrt 73) * (11-sqrt 73)/(11-sqrt 73)
α/(β + 1) = (7+sqrt 73)* (11+sqrt73)/(121-73)
β/(α + 1 ) = (7-sqrt 73) * (11-sqrt 73)/(121- 73)
(77 +18 sqrt 73 +73)/48
(77 - 18 sqrt 73 +73/48
-------------------------------- add
(154 + 146) /48
= 150/48
Answered by
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