# 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

• Your approach is correct. What is the definition of range? I would say range is just $\beta-\alpha$ since $\beta>\alpha$ as you wrote it. – coreyman317 May 19 at 5:28
• I don't see that "range of its roots" has a well-defined meaning. Can you check whether that's exactly how the problem is stated? Or, check previous examples in this context. Otherwise, do they mean ranges of the roots as functions of $k,$ since we always want to have two distinct real roots? Or the range of values of $k$ for which there are distinct roots $\alpha, \beta$? These are things you need to clarify. – Allawonder May 19 at 6:06
• Well range means the set of values which$\alpha$ and $\beta$ can take. And the answer is in the form (a,b). – Abhinav May 19 at 10:54
• You can have an estimation of the ranges of the roots, by finding the location of roots in graph plotted. – Vedant Chourey May 19 at 12:08

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.

• I don't think your answer is a very good solution to my question as well don't draw graphs of such complex equation that too rational equation. – Abhinav May 19 at 10:59
• @Abhinav well if you can't plot a graph then find the value of k where the derivative is 0. And put that value of k in the equation – Shantam Srivastava May 19 at 12:19

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).

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 $$k$$