Continuation of Tutorial?

Here's the other way to solve a quadratic equation. I'll make this short cause I'm tired and I guess- lazy





Here's how it works.





Pretend we have x^2 + 4x + 3 = 0





x^2 + 4x = -3





b = 4





Add by (b/2)^2 on both sides





(4/2)^2 = 4





x^2 + 4x + 4 = 1





Now we can factor this.





Find b/2 and that will be your blank term of:





(x+_)(x+_) = 1





4/2 = 2





(x+2)(x+2) = 1





(x+2)^2 = 1





x + 2 = -1





x + 2 = 1





x = -3 or -1





When you factor normally you would have gotten:





(x+3)(x+1) = 0





x = -3





x = -1





Now pretend we have an equation:





2x^2 + 10x + 12 = 0





You could divide by 2 on both sides, but let's not.





Multiply the a term with the c term.





(2)(12) = 24





Find a factor pair that will add to 10.





The factor pair is 4 and 6.





Split the equation into 2 groups.





(2x^2 + 4x) + (6x + 12) = 0





Factor the insides





2x(x+2) + 6(x+2) = 0





(2x+6)(x+2) = 0





2(x+3)(x+2) = 0





x = -3 or -2





Hope you've learned something today!

Continuation of Tutorial?
Thanks Kenneth you just explained why the famous quadratic formula (-b+-sqrt(b^2-4ac))/2a works. However I don't think there is anything new in what you wrote, people have known about completing to squares and etc for a very long time now and your method is not as fast as using the formula.

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