F.4 Quadratic Equation

α + β = -14/2 = - 7
αβ = - 5/2

1) (α + 1/β) ( β + 1/α)

= αβ + β/β + α/α + 1/αβ

= -5/2 + 1 + 1 + 1/(-5/2)

= - 9/10

2)α^2 + 3αβ + β^2

= α^2 + 2αβ + β^2 + αβ

= (α + β)^2 + αβ

= (-7)^2 + -5/2

= 93/2

3) α/β + β/α

= (α^2 + β^2) / αβ

= [(α + β)^2 - 2αβ ] / αβ

= [(-7)^2 - 2(-5/2)] / (-5/2)

= -108/5

For the equation 2x^2 + 14x - 5 = 0
Sum of roots = α + β = -14/2 = -7
Product of roots = α β = -5/2

So,
1.) (α + 1/β) ( β + 1/α)
= α β + α/α + β/β + (1/αβ)
= (-5/2) +1 +1 (1/(-5/2))
= (-5/2) +2 + (-2/5)
= -9/10

2.) α^2 + 3αβ + β^2
= (α^2 + 2αβ + β^2) + αβ
= (α+β)^2 + αβ
= (-7)^2 + (-5/2)
= 49 - (5/2)
= 93/2

3.) α/β + β/α
= (α^2 + β^2) / αβ
= [(α^2 + 2αβ + β^2) - 2αβ] / αβ
= [(α+β)^2 - 2αβ] / αβ
= [(-7)^2 - 2(-5/2)] / (-5/2)
= (49+5) / (-5/2)
= -108/5


p.s.
When you see this kind of question, you'd better find the sum of roots and products of roots first. Because it mentions that you should find the answer "without solving the equation".
After finding sum and product, substitute into the place that you want.
e.g. product = αβ = (-5/2)
then 3αβ = 3(-5/2)

If α and β are the roots of ax^2 + bx c = 0
then sum of roots = α + β = -b/a
product of roots = αβ = c/a

Also α and β are the roots(roots means solution, the value of x)
then x = α or β
But α is not a
β is not b