$(x^2+2mx+7m-12)(4x^2-4mx+5m-6)=0$ have two distinct real roots

Question

Find the range of $m$ so that the equation $(x^2+2mx+7m-12)(4x^2-4mx+5m-6)=0$ have two distinct real roots.

Either $(x^2+2mx+7m-12)=0$ or $(4x^2-4mx+5m-6)=0$
The discriminant of $(x^2+2mx+7m-12)=0$ is $4(m-3)(m-4)$
and the discriminant of $(4x^2-4mx+5m-6)=0$ is $16(m-2)(m-3)$

I don't know how to do it further.

Answer 1

If $m=3$, then each quadratic has one and only one solution, one of which is $-3$ and the other of which is $\frac32$. So $m=3$ is a solution of your problem.

If $m\in(3,4)$, then the first quadratic has no roots, whereas the second one has exactly two roots. So, each such $m$ is a solution of your problem.

If $m\in(2,3)$, then the first quadratic has exactly two roots, whereas the second one has none. So, each such $m$ is a also solution of your problem.

In every other case, each quadratic has at least one root and at least one of them has two roots.

So, the answer to your problem is $(2,4)$.

Answer 2

Slightly improved:

There are either $(2, 0)$ roots, $(1, 1)$ or $(0, 2)$ in each of the two brackets. This means that the product of the discriminants $64(m-3)^2 (m-2)(m-4)$ must be negative or zero.

This quartic polynomial has a positive leading term and has roots $2, 3, 4$ ($3$ is a double root). Draw a quick sketch:

Hence $2 < m < 4$.

Answer 3

We can also look at the "vertex form" of each quadratic factor of $ \ (x^2+2mx+7m-12)·(4x^2-4mx+5m-6) \ $ to see what they contribute to their product function.

The first factor can be written as $$ f(x) \ \ = \ \ x^2 \ + \ 2mx \ + \ (7m - 12) \ \ = \ \ (x \ + \ m)^2 \ + \ (7m - 12) \ - \ m^2 $$ $$ = \ \ (x \ + \ m)^2 \ - \ (m^2 \ - \ 7m \ + \ 12) \ \ = \ \ (x \ + \ m)^2 \ - \ [ \ (m \ - \ 3)·(m \ - \ 4) \ ] \ \ $$ [as expected, the factors of the discriminant you found appear in the bracketed term of the "completed square"]. The vertex of the "upward-opening" parabola corresponding to this factor is then "above" the $ \ x-$axis when the term in brackets is negative, which occurs for $ \ 3 < m < 4 \ \ . $ We can then say that this factor has

• no real zeroes for $ \ 3 \ < \ m \ < \ 4 \ \ , $

• one real zero at $ \ x \ = \ -m \ $ when $ \ m \ = \ 3 \ $ or $ \ m \ = \ 4 \ \ , \ \ $ and

• two real zeroes for $ \ m \ < \ 3 \ $ or $ \ m \ > \ 4 \ \ . $

Considering the second factor in this way, we find

$$ g(x) \ \ = \ \ 4x^2 \ - \ 4mx \ + \ (5m - 6) \ \ = \ \ 4·\left(x \ - \ \frac{m}{2} \right)^2 \ + \ (5m - 6) \ - \ 4·\left(\frac{m}{2} \right)^2 $$ $$ = \ \ 4·\left(x \ - \ \frac{m}{2} \right)^2 \ - \ (m^2 \ - \ 5m \ + \ 6) \ \ = \ \ 4·\left(x \ - \ \frac{m}{2} \right)^2 \ - \ [ \ (m \ - \ 2)·(m \ - \ 3) \ ] \ \ . $$ The vertex for the "upward-opening" parabola here is again "above" the $ \ x-$axis when the term in brackets is negative for $ \ 2 < m < 3 \ \ . $ So we conclude that this factor has

• no real zeroes for $ \ 2 \ < \ m \ < \ 3 \ \ , $

• one real zero at $ \ x \ = \ \frac{m}{2} \ $ when $ \ m \ = \ 2 \ $ or $ \ m \ = \ 3 \ \ , \ \ $ and

• two real zeroes for $ \ m \ < \ 2 \ $ or $ \ m \ > \ 3 \ \ . $

The quartic polynomial that is the product of these two factors has the number of real zeroes that is the sum of the number of real zeroes in the two factors. Therefore, $ \ f(x) \ · \ g(x) \ $ has, for varying $ \ m \ \ : $

$ \mathbf{m \ < \ 2 \ \ : } \ \ \ 2 + 2 \ = \ 4 \ $ real zeroes;

$ \ \ \mathbf{m \ = \ 2 \ \ : } \ \ \ 2 + 1 \ = \ 3 \ $ real zeroes (one of them at $ \ x \ = \ \frac{m}{2} \ = \ \frac{2}{2} \ = \ +1 \ \ ) \ ; $

$ \mathbf{2 \ < \ m \ < \ 3 \ \ : } \ \ \ 2 + 0 \ = \ 2 \ $ real zeroes;

$ \ \ \mathbf{m \ = \ 3 \ \ : } \ \ \ 1 + 1 \ = \ 2 \ $ real zeroes (one of them at $ \ x \ = \ -m \ = \ -3 \ \ , $ the other at $ \ x \ = \ \frac{m}{2} \ = \ +\frac{3}{2} \ \ ) \ ; $

$ \mathbf{3 \ < \ m \ < \ 4 \ \ : } \ \ \ 0 + 2 \ = \ 2 \ $ real zeroes;

$ \ \ \mathbf{m \ = \ 4 \ \ : } \ \ \ 1 + 2 \ = \ 3 \ $ real zeroes (one of them at $ \ x \ = \ -m \ = \ = \ -4 \ \ ) \ ; $

and $ \ \ \mathbf{ m \ > \ 4 \ \ : } \ \ \ 2 + 2 \ = \ 4 \ $ real zeroes.

Hence, the quartic polynomial has two distinct real zeroes for $ \ \mathbf{2 \ < \ m \ < \ 4} \ \ ; \ \ $ these are the zeroes of either factor alone or from both vertices at once $ \ ( \ m \ = \ 3 \ ) \ \ . $

The polynomial never has less than two real zeroes for any value of $ \ m \ \ . \ $ The third zero for $ \ m \ = \ 2 \ $ and $ \ m \ = \ 4 \ $ must be a "double zero"; the two zeros for $ \ m \ = \ 3 \ $ are both "double zeroes". In fact, this polynomial never has complex zeroes.

[There are places where one could be deceived by using a low-resolution graph or not examining a graph carefully enough. The function $ \ f(x) \ · \ g(x) \ $ has a turning point extremely close to the $ \ x-$axis for

$ m \ \approx \ -2.25 \ $ near $ \ x \ \approx \ -3.482 \ \ $ (slightly below); $ \ m \ \approx \ +1.12 \ $ near $ \ x \ \approx \ +1.205 \ \ $ (slightly below); and $ \ m \ \approx \ +1.53 \ $ near $ \ x \ \approx \ +0.362 \ \ $ (slightly above). ]