Question

Use completing the square to solve: 3x^2+12x-9=0

Answer 1

First, simplify,
\[3(x^2+4x-3)=0\Rightarrow x^2+4x-3=0\]Now, complete the squares, using
\[(a+b)^2=a^2+b^2+2ab\]You have,
\[4x=2\cdot x\cdot b \Rightarrow b=2\]
\[x^2+4x-3=x^2+4x+2^2-2^2-3=(x+2)^2-7=0\Rightarrow x+2=\pm\sqrt{7}\Rightarrow\\
\Rightarrow x=-2\pm\sqrt{7}\]

Answer 2

what happened after the 2nd step?

Answer 3

@John_ES

Answer 4

You have a term 2ab, that you need to find in your equations. As there is only one term that has a number multiplied by an x (not x^2), then you can use the 3 step.

Answer 5

The term -3 is out of the perfect square as its sign is negative.

Answer 6

oh ok