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5 votes
Does anyone know how to do this?

Does anyone know how to do this?-example-1
User Ajo
by
6.5k points

2 Answers

0 votes
It is factorable if you can solve it when it equals 0.
There are many ways to do this.

X^2+bx+12=0


(-b +- \sqrt{ b^(2) -4*12 } )/2

This equation gives a good answer if

b ^(2) \geq 48
User Joekarl
by
6.7k points
3 votes
You're looking for values to factor this equation into (x+p)(x+q)
such that pq=12 and p+q=b

Assuming we're looking for simple values, you can start by factorizing 12:

12 = 3*4 = 2*6 = 12*1 = -3*-4 = -2*-6 = -12*-1

For 3 and 4, b would be 7
2 and 6 => b=8
12 and 1 => b=13

and for the negatives

b can be -7, -8 or -13

So the total set is -13, -8, -7, 7, 8, 13
User Jason Wiener
by
6.6k points
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