Let
4x^2-4(α-2)x+α-2=0
(α is real number)
be a quadratic equation, then find the value of α for which both the roots are positive.
the conditions will be
1) Discriminant D≥0
by this condition i got α (-∞,2][3,∞)
2) f(0) greater than or equal to 0
by this we get α (2,∞)
3) now should i use -b/2a(point exactly between both roots)
and equate as -b/2a greater than 0
if-3rd point is right then what will be the final answer
α (?,?)union(?,?)
please provide help
What you've done is fine. The condition that the equation has two roots can be satisfied by and so
To continue I would say, let be the two positive roots of the function and therefore:
Now, You have to solve two irrational inequality's in function of and so you'll find out the values of wherefore and
This is the way I would try to solve this question.
Thanks i got the answer,
But in a different way:-
This method is based on the graph of quadratic equation that is a parabola
In third point ' 3)...... '
As (-b/2a ,-D/4a) is the vertex of parabola
it's x component lies exactly in the middle of the two roots( in case of one roots they coincide)
So it(-b/2a) will be greater than the smallest root,
and hence it will be greater than zero.
As zero is the point on x axis from where positive numbers start,
so taking it as reference point.
We get that -b/2a will be greater than 0
equating this we will get an interval (2,∞).
From this and other two interval in question
(-∞,2][3,∞)-------------(1)
(2,∞) ------------(2)
We get final interval as
[3,∞)
Thank you for your method.To continue I would say, let \delta, \gamma be the two positive roots of the function and therefore:
\delta=\frac{-b+\sqrt{D}}{2a}=\frac{4(\alpha-2)+\sqrt{16(\alpha-2)^2-16(\alpha-2)}}{8}\geq 0
\gamma =\frac{-b-\sqrt{D}}{2a}=\frac{4(\alpha-2)-\sqrt{16(\alpha-2)^2-16(\alpha-2)}}{8}\geq 0
Now, You have to solve two irrational inequality's in function of \alpha and so you'll find out the values of \alpha wherefore \delta\geq 0 and \gamma \geq 0
I will try this