Completing the square with second degree coefficient greater than one

Question

How do I complete the square when the second degree coefficient is greater than one. I can do it when $x^2+4x-4=0$, for example, but I can't work out how to do when $3x^2+4x-4=0$.

Answer 1

Best solution is to normalize the leading coefficient. I.E. $$3x^2+4x-4=0 \implies x^2 +\frac{4x}{3}-\frac{4}{3}=0 \implies (x^2+\frac{4x}{3}+ ~~~)-\frac{4}{3}=0$$ We know that for completing the square, we should have the $c$ term equal to the square of half the $b$ term divided by the $a$ term. $c=\left(\frac{b}{2a}\right)^2$ $$ (x^2+\frac{4x}{3}+ \frac{4}{9})-\frac{4}{9}-\frac{4}{3}=0 \implies \left(x+\frac{2}{3}\right)^2-\frac{16}{9}=0$$ Solve: $$\left(x+\frac{2}{3}\right)^2=\frac{16}{9} \implies \left( x+\frac{2}{3}\right)=\pm\frac{4}{3}$$ Therfore: $$x=\frac{2}{3},-2$$

Answer 2

Hint: $$3x^2+4x-4 = 3\left(x^2+\frac{4}{3}x-\frac{4}{3}\right).$$

Answer 3

$$3x^2+4x-2=3\left(x^2+\frac43x-\frac43\right)=3\left[\left(x+\frac23\right)^2-\frac49-\frac43\right]=\ldots$$

Answer 4

If you would prefer to avoid fractions as long as possible multiply by four times the leading coefficient:

$$12(3x^2+4x-4)=36x^2+48x-48=(6x+4)^2-64$$ In general $$4a(ax^2+bx+c)=4a^2x^2+4abx+4ac=(2ax+b)^2+4ac-b^2$$