Find the interval in which $a$ lies such that the roots of the equation $(a+1)x^2 - 3ax + 4a$, is greater than 1.

Question

Question: Find the interval in which $a$ lies such that the roots of the given function, $f(x)$, is greater than 1. $$f(x) = (a+1)x^2 - 3ax + 4a$$

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My Attempt:

Using the quadratic formula, $$x = \frac{3a±\sqrt{9(a)^2-16(a)^2-16a}}{2(a+1)}>1$$

First of all, the radicand must be greater than or equal to $0$. From that, we can restrict $a$ to $$a \in [-\frac{16}{7}, 0]$$

Here, we can make two cases:

1. Case I: $a+1>0$

$$±\sqrt{9(a)^2-16(a)^2-16a}>2 - a$$

Now squaring both sides and simplifying it further,

$$-7(a)^2-16a >a^2 + 4 -4a$$ $$2(a)^2+3a+1 <0$$ Hence, $a \in (-1, -\frac{1}{2})$

Intersecting this with the restricted set, we get $$a \in (-1, -\frac{1}{2})$$

2. Case II: $a+1 < 0$

Here, we will end up getting, $$a \in [-\frac{16}{7}, -1)$$

But this is not the required answer. Using desmos, it can be clearly seen that for $a\in(-1, -\frac{1}{2})$ there is only one root greater than $1$ and not both.

I suspect that I messed it up while squaring the radical. By squaring, I was trying to find the interval for $a$ where at least one root has a value greater than $1$. Right?

How am I supposed to get around it? Also, is there a better approach for this particular problem?

Answer 1

Note that solving $\frac{3a}{2(a+1)}<1$ gets you to (-1,2). In this region the - solution cannot be bigger than 1 since you are subtracting a positive number, so you have to exclude that region

Answer 2

To find the values of $a$ such that real roots of the equation $$ (a+1)x^2 - 3ax + 4a=0$$ are greater than $1$.

First the discriminant must be positive $$(3a)^2-16a(a+1)\ge 0\to -\frac{16}{7}\leq a\leq 0\tag{1}$$

Second, if we want both roots greater than $1$, then the sum must be greater than $2$

The sum of the roots is $x_1+x_2=\frac{3a}{a+1}$

$$x_1+x_2>2\to \frac{3a}{a+1}>2\to a<-1\lor a>2\tag{2}$$

Putting together $(1)$ and $(2)$ we have that the roots of the given equation are greater than $1$ if $-\frac{16}{7}\leq a<-1$