# Thread: Location of roots problem.

1. ## Location of roots problem.

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(?,?)

2. ## Re: Location of roots problem.

Originally Posted by sumedh
Let $4x^2-4(α-2)xα-2=0 (α\epsilon R)$
What does this mean?

3. ## Re: Location of roots problem.

sorry the alpha are not seen α=alpha
4x^2-4(α-2)x+α-2=0
where alpha is real.

4. ## Re: Location of roots problem.

What you've done is fine. The condition that the equation has two roots can be satisfied by $D\geq 0$ and so
$\alpha \in ]-\infty, 2]\cup [3,+\infty[$

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 $\gamma \geq 0$

This is the way I would try to solve this question.

5. ## Re: Location of roots problem.

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,∞)

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

6. ## Re: Location of roots problem.

Originally Posted by sumedh
Let

$4x^2-4(a-2)x+a-2=0$

(a is real number)

be a quadratic equation, then find the value of $a$ for which both the roots are positive.

the conditions will be
1) Discriminant $D\ge 0$ by this condition i got $a$: $(-\infty,2][3,\infty)$
Now Descartes rule of signs tells us under the conditions in force here that neither of the roots are negative iff $(a-2)\ge 0$, or rather $a\ge 2$, so we are now restricted to $a \in [3,\infty)$.

CB

7. ## Re: Location of roots problem.

Thank you very much