MHB Solving Quadratic Equations: m^4+4m^3+8m^2+8m+4=0

[danny12345]

m^4+4m^3+8m^2+8m+4=0

 

[MarkFL]

Using the rational roots theorem, do you find any rational roots?

That is, let:

$$P(m)=m^4+4m^3+8m^2+8m+4$$

Obviously, if there are any rational roots, they must be negative since each term has a plus before it. So, check the following:

$$m\in\{-1,-2,-4\}$$

And see if any of those value for $m$ makes $P=0$. What do you find?

 

[danny12345]

none of them gave 0.

 

[MarkFL]

none of them gave 0.

Yes, so that means there are no rational roots. The next thing we can try is to see if the quartic polynomial can be factored into two quadratics. So, I would write:

$$m^4+4m^3+8m^2+8m+4=\left(m^2+am+b\right)\left(m^2+cm+d\right)$$

Expand the right side, and then equate corresponding coefficients to get a 4X4 system of equations. What do you find?

 

[danny12345]

solve it for me

 

[MarkFL]

solve it for me

You need to be able to do things like this for yourself, otherwise you will encounter great difficulty down the road. Expanding the right side can be made simpler by writing:

$$\left(m^2+am+b\right)\left(m^2+cm+d\right)=m^2\left(m^2+cm+d\right)+am\left(m^2+cm+d\right)+b\left(m^2+cm+d\right)$$

Now distribute in each of the 3 terms, and them combine like terms (those terms that have the same exponent on $m$)...what do you get.

I will be glad to guide you, but I won't just work the problem for you. Not because I am a big ol' meanie, but because I want you taking part in the process, getting dirt under your fingernails and coming through this with a better understanding than you would get if you just see it solved.

When my dad taught me to change an automobile tire when I was a kid, he didn't say, "Son, grab a chair and watch me change this tire." He said, "Son, grab the jack and tire iron from the trunk, and I will teach you how to change a tire." :)

 

[danny12345]

m^4+m^2(d+ac+b)+m^3(c+a)+m(ad+bc)+bd=0

- - - Updated - - -

m^4+m^2(d+ac+b)+m^3(c+a)+m(ad+bc)+bd=0
now how do iget the coefficient value

 

[MarkFL]

m^4+m^2(d+ac+b)+m^3(c+a)+m(ad+bc)+bd=0

- - - Updated - - -

m^4+m^2(d+ac+b)+m^3(c+a)+m(ad+bc)+bd=0
now how do iget the coefficient value

Yes, so what we have now is:

$$m^4+4m^3+8m^2+8m+4=m^4+(a+c)m^3+(d+ac+b)m^2+(ad+bc)m+bd$$

So, equating coefficients, we obtain the system:

$$a+c=4$$

$$d+ac+b=8$$

$$ad+bc=8$$

$$bd=4$$

Can you perhaps spot an easy solution?

 

[Greg]

You end up with the following equations:

$$a+c=4$$

$$b+ac+d=8$$

$$ac+bc=8$$

$$bd=4$$

Not an easy system to solve, I dare say.

However, note that $$bd=4$$. This gives ordered pairs $$(-1,-4),(-2,-2),(2,2),(1,4)$$ for $$b$$ and $$d$$ (not necessarily in that order). By checking these pairs one can indeed determine that the given quartic is bi-quadratic (the product of two quadratics).

 

[anemone]

For me, I'd do it as follows:

I'd first factor $m^4+4m^3+4m^2$ in $m^4+4m^3+8m^2+8m+4=0$, it's just a hunch, or my experience of solving equations that prompted me to do that:

$\begin{align*}m^4+4m^3+8m^2+8m+4&=m^4+4m^3+4m^2+4m^2+8m+4\\&=m^2(m^2+4m+4)+4m^2+8m+4\\&=m^2(m+2)^2+4m^2+8m+4\\&=(m^2+2m)^2+4(m^2+2m)+4\color{green}\,\,\,\text{compare it with}=a^2+4a+4=(a+2)^2\color{black}\\&=(m^2+2m+2)^2\end{align*}$

 

[kaliprasad]

I see the pattern as
$m^4+4= (m^2+2)^2 - 4m^2$
and
$4m^3+8m = 4m(m^2+2)^2$
so we see $m^2+2$ as common and we proceed as
$m^4+4m^3+8m^2+8m+4$
= $(m^4 + 4)+ (4m^3 + 8m) + 8m^2$
= $(m^2 + 2)^2 - 4m^2 + 4m(m^2 + 2) +8m^2$
= $(m^2+2)^2 + 4m(m+2) + 8m^2$
= $(m^2+2)^2 + 2 * (2m(m+2)) + (2m)^2$
= $(m^2 + 2 + 2m)^2$

another problem was solved in this method also http://mathhelpboards.com/challenge-questions-puzzles-28/find-sum-real-roots-7882.html

 

[Greg]

$$u=m^2+2,u^2=(m^2+2)^2=m^4+4m^2+4$$

$$m^4+4m^3+8m^2+8m+4=(m^2+2)^2+4m(m^2+2)+4m^2$$

$$=(u+2m)^2=(m^2+2m+2)^2$$