If α and β are the root of the equation 2X²-7X-3=0 Find
α
---- + 1 +
β
β
---- + 1
α
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 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
To find the value α + 1/β + β/α + 1, we first need to find the values of α and β, which are the roots of the equation 2X² - 7X - 3 = 0.
To solve this quadratic equation, we can use the quadratic formula, which states that for an equation of the form AX² + BX + C = 0, the roots can be found using the formula:
X = (-B ± √(B² - 4AC)) / 2A
In our equation, A = 2, B = -7, and C = -3. Plugging these values into the quadratic formula, we can find the roots α and β.
For α:
X = (-(-7) ± √((-7)² - 4(2)(-3))) / (2(2))
= (7 ± √(49 + 24)) / 4
= (7 ± √73) / 4
Therefore, α = (7 + √73) / 4 and/or α = (7 - √73) / 4.
Similarly, for β:
X = (-(-7) ± √((-7)² - 4(2)(-3))) / (2(2))
= (7 ± √(49 + 24)) / 4
= (7 ± √73) / 4
Therefore, β = (7 + √73) / 4 and/or β = (7 - √73) / 4.
Now that we have the values of α and β, we can substitute them into the expression α + 1/β + β/α + 1:
(α + 1/β) + (β/α + 1)
[(αβ + 1) + β(1 + α)] / αβ
[αβ + 1 + β + αβ] / αβ
[2αβ + 1 + α + β] / αβ
Substituting the values of α and β:
[2((7 + √73) / 4)((7 - √73) / 4) + 1 + (7 + √73) / 4 + (7 - √73) / 4] / ((7 + √73) / 4)((7 - √73) / 4)
Simplifying this expression would result in a numerical value.