Question

What are the zeros of the equation x^2+4x-12=0

Answer 1

The first method:

`x^2+4x-12=0\ \ \ |+12\\\\x^2+2\cdot x\cdot2=12\ \ \ \ |+2^2\\\\\underbrace{x^2+2\cdot x\cdot2+2^2}_((a+b)^2=a^2+2ab+b^2)=12+2^2\\\\(x+2)^2=16\to x+2=\pm√(16)\\\\x+2=-4\ \vee\ x+2=4\ \ \ |-2\\\\x=-6\ \vee\ x=2`

The second method:

`x^2+4x-12=0\\\\a=1;\ b=4;\ c=-12\\\\b^2-4ac=4^2-4\cdot1\cdot(-12)=16+48=64\\\\√(b^2-4ac)=√(64)=8\\\\x_1=(-b-√(b^2-4ac))/(2a);\ x_2=(-b+√(b^2-4ac))/(2a)\\\\x_1=(-4-8)/(2\cdot1)=(-12)/(2)=-6\\\\x_2=(-4+8)/(2\cdot1)=(4)/(2)=2`

Third method:

`x^2+4x-12=0\\\\x^2+6x-2x-12=0\\\\x(x+6)-2(x+6)=0\\\\(x+6)(x-2)=0\iff x+6=0\ \vee\ x-2=0\\\\x=-6\ \vee\ x=2`