MHB Finding the roots of quadratic equation

[paulmdrdo1]

can you show me a way of solving this problem without considering the discriminant.

Find the roots of equation subject to the given condition.

$(2m + 1)x^2-4mx = 1-3m$ has equal roots.

I solved it using discriminant but I want to know other way of solving it. Thanks!

 

[kaliprasad]

can you show me a way of solving this problem without considering the discriminant.

Find the roots of equation subject to the given condition.

$(2m + 1)x^2-4mx = 1-3m$ has equal roots.

I solved it using discriminant but I want to know other way of solving it. Thanks!

you can complete the square and make the constant part as zero

first multiply by (2m + 1) to avoid redical

$((2m + 1)x)^2-2(2m)((2m+1)x) = (1-3m)(2m+1)$

add 4m^2 on both sides to complete square
$((2m + 1)x)^2-2(2m)((2m+1)x) + (2m)^2 = (1-3m)(2m+1)+ 4m^2$

or $((2m+1)x- 2m)^2 = (1-3m)(2m+1)+ 4m^2$

for the roots to be equal RHS has to be zero

now you should be above to proceed

 

[Deveno]

Another way to do this is by "comparing coefficients".

If:

$x^2 - \dfrac{4m}{2m+1}x + \dfrac{3m - 1}{2m+1} = 0 = (x-r)^2 = x^2 - 2rx + r^2$

we get two equations:

$2r = \dfrac{4m}{2m+1}$

$r^2 = \dfrac{3m - 1}{2m + 1}$.

Thus:

$r^2 = \left(\dfrac{2m}{2m+1}\right)^2 = \dfrac{3m-1}{2m+1}$.

This allows you to solve for $m$, and thus solve for $x = r$.

 

[paulmdrdo1]

where did you get this?

$(x-r)^2 = x^2 - 2rx + r^2$

I thought something similar to this by letting r to be the roots of my equation.

since $S = \frac{-b}{a}=r+r$

I get $\frac{-b}{a}=2r$

and $P=\frac{c}{a}=r\times r$

I get $ \frac{c}{a}=r^2$

now I will have the same equation as you do

$2r=\frac{4m}{2m+1}$

and

$r^2=\frac{3m-1}{2m+1}$

Am I thinking it the same way that you did?

the r here is the value that I will plug in place of x in my equation.

but from this $(x-r)^2 = x^2 - 2rx + r^2$ It seems x and r are different?

why is that? thanks!

 

[Deveno]

In THIS equation:

$(x-r)^2 = x^2 - 2rx + r^2$

$r$ and $x$ ARE different.

But in THIS equation:

$(x - r)^2 = 0$

we must have $x - r = 0$ (0 is the square root of 0), and thus $x = r$.

And yes, you were thinking correctly, good job!

 

[paulmdrdo1]

Oh yes that's helpful. :) thanks!