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It was hard for me to factorize Quadratic Equations and I read many explanations. I come up with two simple examples that , from my point of view, are good templates to solve the majority of Quadratic Equations.
Simple case (no number before \(x^2\)):
\(x^2\)+Bx+C=0
Solution:
\(x_1\)+\(x_2\)=-B
and
\(x_1\)*\(x_2\)=C
two numbers that works are the \(x_1\) and \(x_2\) values.
In case you need to write in an equation, It will look this way (x-\(x_1\))*(x-\(x_2\))=0
Now normal example,
\(x^2\)+5x+6=0
We need:
\(x_1\)+\(x_2\)=-5
and
\(x_1\)*\(x_2\)=6
So, \(x_1\) =-2 and \(x_2\)=-3
(x+2)*(x+3)=0
Hard case (there is a number before \(x^2\)):
A\(x^2\)+Bx+C=0
Solution:
\(x_1\)+\(x_2\)=B
and
\(x_1\)*\(x_2\)=[color=C*A
two numbers that works are the \(x_1\) and \(x_2\) values.
In case you need to write in an equation, It will look this way \(A(x+\frac{x_1}{A})*(x+\frac{x_2}{A})=0\)
Now normal example,
2\(x^2\)+x-6=0
We need:
\(x_1\)+\(x_2\)=1
and
\(x_1\)*\(x_2\)=2*-6=-12
So, \(x_1\) =4 and \(x_2\)=-3
\(2(x+\frac{4}{2})*(x-\frac{3}{2})=0\)
Hope my post helps someone,
If it does pls, dont hesitate to give Kudos.
Simple case (no number before \(x^2\)):
\(x^2\)+Bx+C=0
Solution:
\(x_1\)+\(x_2\)=-B
and
\(x_1\)*\(x_2\)=C
two numbers that works are the \(x_1\) and \(x_2\) values.
In case you need to write in an equation, It will look this way (x-\(x_1\))*(x-\(x_2\))=0
Now normal example,
\(x^2\)+5x+6=0
We need:
\(x_1\)+\(x_2\)=-5
and
\(x_1\)*\(x_2\)=6
So, \(x_1\) =-2 and \(x_2\)=-3
(x+2)*(x+3)=0
Hard case (there is a number before \(x^2\)):
A\(x^2\)+Bx+C=0
Solution:
\(x_1\)+\(x_2\)=B
and
\(x_1\)*\(x_2\)=[color=C*A
two numbers that works are the \(x_1\) and \(x_2\) values.
In case you need to write in an equation, It will look this way \(A(x+\frac{x_1}{A})*(x+\frac{x_2}{A})=0\)
Now normal example,
2\(x^2\)+x-6=0
We need:
\(x_1\)+\(x_2\)=1
and
\(x_1\)*\(x_2\)=2*-6=-12
So, \(x_1\) =4 and \(x_2\)=-3
\(2(x+\frac{4}{2})*(x-\frac{3}{2})=0\)
Hope my post helps someone,
If it does pls, dont hesitate to give Kudos.
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