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?

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

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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