Find all possible values $m$ such that $f(x)=x^2-5mx+10m-4$ has 2 roots and one of them is twice as the other

Question

Consider the polynomial $f(x)=x^2-5mx+10m-4$, find $m$ such that there exist a number $a$ that satisfies $f(a)=f(2a)=0$. This was my attempt:

$f(a)=f(2a)$
$a^2-5ma+10m-4=4a^2-10ma+10m-4$
$a^2-5ma=4a^2-10ma$
$-3a^2+5ma=0$
$a*(-3a+5m)=0$

If $a=0$ then $f(a)=10m-4=0$ and $m=\frac{2}{5}$, else if $-3+5m=0$ then I can't solve.

Is my first answer correct? And if yes, how do I solve for the second one

Answer 1

Your equation always has two roots: $2$ and $5m-2$. So, one of them is twice the other if and only if $m=\frac65$ or if $m=\frac35$.

In your approach, it seems that you forgot that you should also have $f(a)=0$.

Answer 2

The given equation can be written as

$\begin{align} f(x)& =x^2-4-5mx+10m\\ &=(x-2) (x+2) -5m(x-2) \\ &=(x-2) (x+2-5m) \end{align}$

Now, we have two possibilities: $\alpha=2\beta$ or $\beta=2\alpha$.

Answer 3

Let the roots be $\alpha, 2\alpha$.

By Vieta's formulas,

Sum of roots $=3\alpha = 5m$, product of roots $=2\alpha^2 = 10m-4$.

Hence, $2(\frac{5m}3)^2 = 10m-4 \\ \implies 25m^2 - 45m +18 = 0$, we get (by simple factorisation) $m=0.6$ or $1.2$.

Test by plugging these back into the original expression and getting $f(x) = x^2-3x+2$ and $f(x) = x^2 - 6x+8$ respectively, which each have two distinct roots with one twice the other.

Answer 4

Another approach without making use of the quadratic formula is to apply the Factor Theorem. If the zeroes of $ \ x^2 - 5mx + (10m-4) \ $ are $ \ x = a \ $ and $ \ x = 2a \ \ , $ then we can write the polynomial as $$ x^2 \ - \ 5mx \ + \ (10m - 4) \ \ = \ \ ( x - a ) · (x - 2a) \ \ = \ \ x^2 \ - \ 3ax \ + \ 2a^2 \ \ . $$

Making the correspondences between the linear and constant terms of the two expressions gives us $ \ -3a \ = \ -5m \ $ and $ \ 2a^2 \ = \ 10m - 4 \ \ . $ (This avoids the problem of needing to "see" how to factor the given polynomial. This also makes it evident how the Viete relations in Deepak's answer arise.) The first of these equations tells us that $ \ m \ = \ \frac35·a \ \ , $ which we can then insert into the second equation to obtain $$ 2a^2 \ - \ 10m \ + \ 4 \ \ = \ \ 0 \ \ \Rightarrow \ \ a^2 \ - \ 5·\left(\frac35·a \right) \ + \ 2 \ \ = \ \ 0 $$ $$ \Rightarrow \ \ a^2 \ - \ 3a \ + \ 2 \ \ = \ \ (a \ - \ 1) · (a \ - \ 2) \ \ = \ \ 0 \ \ . $$

We find that either $ \ a = 1 \ $ or $ \ a = 2 \ \ , $ which makes $ \ 2a \ $ equal to either $ \ 2 \ $ or $ \ 4 \ \ . $ (This agrees with conclusions in other posted answers that one of the zeroes of the polynomial must be $ \ 2 \ \ . ) $ Returning to the relation $ \ m \ = \ \frac35·a \ \ , $ the only possible values for $ \ m \ $ are thus $ \ \frac35 \ $ or $ \ \frac65 \ \ . $