Number of ways to factorize/solve

Dec 2015
42
1
Europe
Hello! I'm curious to find out how many ways there is to find a solution to the quadratic equation:

\(\displaystyle x^2+5x+6=0\)


I have three ways, do you guys have any more neat tricks?

1)

I factorize in the following way:
\(\displaystyle x^2+5x+6=(x^2+3x)+(2x+6)=x(x+3)+2(x+3)=(x+2)(x+3)\)
So the equation has the solutions \(\displaystyle x=-2 \vee x=-3\)

2)
I use the quadratic formula and plug in the respective values:
\(\displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

and get the solutions \(\displaystyle x=-2 \vee x=-3\)

3) I use the rational root theorem and check for which values \(\displaystyle k \mid 6\) and check where \(\displaystyle k \in {-1,1,-2,2,-3,3,6,6}\) and get that for \(\displaystyle f(-2)=0\) and \(\displaystyle f(-3)=0\)


This is all I can come up with; what about you guys?
 
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skeeter

Math Team
Jul 2011
3,196
1,724
Texas
Complete the square ...

$x^2+5x=-6$

$x^2+5x+\dfrac{25}{4} = -6+\dfrac{25}{4}$

$\left(x+\dfrac{5}{2}\right)^2 = \dfrac{1}{4}$

$x+\dfrac{5}{2} = \pm \dfrac{1}{2}$

$x = -\dfrac{5}{2} \pm \dfrac{1}{2}$

$x=-2$, $x=-3$

Of course, the quadratic formula is derived by completing the square.
 
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Dec 2015
42
1
Europe
Complete the square ...

$x^2+5x=-6$

$x^2+5x+\dfrac{25}{4} = -6+\dfrac{25}{4}$

$\left(x+\dfrac{5}{2}\right)^2 = \dfrac{1}{4}$

$x+\dfrac{5}{2} = \pm \dfrac{1}{2}$

$x = -\dfrac{5}{2} \pm \dfrac{1}{2}$

$x=-2$, $x=-3$

Of course, the quadratic formula is derived by completing the square.
Yes, this is essentially the same thing as using the quadratic formula. Do you have any different methods of attacking this "problem"?
 
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Dec 2015
42
1
Europe
I found one new way:

4) It is essentially the same as 3) because you can use the rational root theorem to find a factor and then continue with Polynomial long division to find the remaining factor.
 
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skeeter

Math Team
Jul 2011
3,196
1,724
Texas
I found one new way:


4) It is essentially the same as 3) because you can use the rational root theorem to find a factor and then continue with Polynomial long division to find the remaining factor
note that the majority of polynomial equations with rational solutions are "made up" to work out nice ... a "math teacher" thing to teach factoring and the rational root theorem.

most real-world applications involving polynomial equations result in irrational solutions.
 
Jun 2016
14
0
New York
Of course, the quadratic formula is derived by completing the square.
Stupid question, but how is the quadratic formula derived from completing the square? I was never aware of this.
 
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Azzajazz

Math Team
Nov 2014
690
245
Australia
Mimic skeeter's first post to solve the equation $ax^2 + bx + c = 0$. The first step is to divide both sides by $a$.
 
Dec 2015
42
1
Europe
$ax^2+bx+c=0$

We subtract $c$and divide $a$ on both sides

$x^2+\frac{b}{a}x=-\frac{c}{a}$

We add $(\frac{b}{2a})^2$on both sides

$x^2+\frac{b}{a}x+(\frac{b}{2a})^2=(\frac{b}{2a})^2-\frac{c}{a}$

We simply left side using the first squaring binomials, and multiply
$-\frac{c}{a}$ with $4a$

$(x+\frac{b}{2a})^2=(\frac{b}{2a})^2-\frac{4ac}{4a^2}$

We take the square root on both sides

$x+\frac{b}{2a}=\pm\sqrt{\frac{b^2}{4a^2}-\frac{4ac}{4a^2}}$

We subtract $\frac{b}{2a}$ on both sides

$x=-\frac{b}{2a}\pm\sqrt{\frac{b^2}{4a^2}-\frac{4ac}{4a^2}}$

We simplify:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
 
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