# Math Help - Equivalence problem

1. ## Equivalence problem

I have to find with what value ''m'' those two equations are equivalent?

$x^2$ + $m^2$x + (6-5m)x = 0

and $x^2$ +2(m-3)x + $m^2$ -7m +12 = 0

I found that with m=3 ... I did this: $x^2$ + $m^2$x + (6-5m)x = $x^2$ +2(m-3)x + $m^2$ -7m +12

Not sure if this is the right way?

2. Originally Posted by Ellla
I have to find with what value ''m'' those two equations are equivalent?

$x^2$ + $m^2$x + (6-5m)x = 0

and $x^2$ +2(m-3)x + $m^2$ -7m +12 = 0

I found that with m=3 ... I did this: $x^2$ + $m^2$x + (6-5m)x = $x^2$ +2(m-3)x + $m^2$ -7m +12

Not sure if this is the right way?
What if m=4 ?

You should write something like

$x^2+\left(m^2-5m+6\right)x+0=0$

$x^2+2(m-3)x+\left(m^2-7m+12)=0$

and comparing, we solve for

$m^2-7m+12=0$

3. Originally Posted by Archie Meade
What if m=4 ?

You should write something like

$x^2+\left(m^2-5m+6\right)x+0=0$

x^2+2(m-3)x+\left(m^2-7m+12)=0

and comparing, we solve for

m^2-7m+12=0
So we get m = 2,3 and 4. If I got it right.

4. m=2 doesn't work.

$m^2-5m+6=2(m-3)$

as the multipliers of "x" must be equal.

$m^2-5m+6=2m-6$

$m^2-5m-2m+6+6=2m-2m-6+6$

$m^2-7m+12=0$

which is the exact same as the two constants being equal
(the constant in one equation is zero).

Then to find the 2 solutions for "m",
find the factors of 12 that sum to -7
in order to factor the quadratic in "m".