To find Range of roots of quadratic equation I have been given the following quadratic equation and is asked to find the range of its roots $\alpha$ and $\beta$, where $\alpha>\beta$
$$(k+1)x^2 - (20k+14)x + 91k +40 =0,$$ 
where $k>0$ .
Here's my approach.

I applied the quadratic formula for the roots and got.
$$\alpha=\frac{(10k+7) -3\sqrt{k^2+k+1}}{k+1}$$
Similarly
$$\beta=\frac{(10k+7)+3\sqrt{k^2+k+1}}{k+1}$$
But how to find the range. Please help
 A: First, we need to find for what values of k such that the equation has real roots ($\alpha$ and $\beta$).
To this end, we set $(10k + 7)^2 \ge (k+1)((91k + 40)$
This is reduced to $k^2 + k + 1 \ge 0$.
The LHS has no real roots for k. In addition, $((k^2))$ is positive. That means the quadratic expression in k is positive definite (i.e. always bigger than 0). 
Therefore, the given equation has real roots for all values of k (except possibly when k = -1). 
A: We know from the question that  $ \alpha > \beta $ ,using the Quadratic formula, the root with the positive sign will be $ \alpha $ and 
the one with the negative sign will be $ \beta $, which you have reversed in your
description.
$ \alpha = \frac{10k+7 + 3\sqrt{k^2 + k + 1}}{k + 1}$
and the other root will be $\beta$
After that, plot the graph of $ \alpha $ versus k, with k as the x-axis. 
The maximum value of that function for k > 0, will be the upper limit of
the range of the root.
Similarly, plot $ \beta $ versus k, with k as the x-axis. 
The minimum value of that function for k > 0, will be the lower limit of the
range of the root. 
A: Your equation can be re-written as $$f(x)=(x-4)(x-10)+k(x-7)(x-13)$$
So it can be seen that
$f(4)=27k$
$f(7)=-9$
$f(10)=-9k$
$f(13)=27$
Since $f(7)$ and $f(13)$ are of opposite signs,clearly one root must be lying on the interval $(7,13)$. 
Therefore we can say that another real root must also exist since in a quadratic equation with real coefficients complex/imaginary roots can only occur in conjugate pairs.
Hence the equation will have real roots for all real values of $k$
