How do I complete the square if $f (x) = 3 x^{2} + 12 x - 9$ $f(x) = 3x^2 + 12x - 9$ ?

Question

11226 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-09-13T22:42:17

Begin by adding 9 to both sides of the equation.

$3 x^{2} + 12 x = 9$ $3x^2+12x=9$

Factor out a $3$ $3$ from the left side of the equation.

$3 (x^{2} + 4 x) = 9$ $3(x^2+4x)=9$

Locate the $x$ $x$ term and take half of the coefficient and square it

${(\frac{4}{2})}^{2} = 2^{2} = 4$ $(4/2)^2=2^2=4$

Add $3 \cdot 4$ $3*4$ to the right side of the equation. Because we factored out a $3$ $3$ .

On the left side add $4$ $4$ within the parenthesis.

$3 (x^{2} + 4 x + 4) = 9 + 12$ $3(x^2+4x+4)=9+12$ , Perfect square trinomial

$3 {(x + 2)}^{2} = 9 + 12$ $3(x+2)^2=9+12$

$3 {(x + 2)}^{2} = 21$ $3(x+2)^2=21$

${(x + 2)}^{2} = 7$ $(x+2)^2=7$ , Divide by 3

$\sqrt{{(x + 2)}^{2}} = \sqrt{7}$ $sqrt((x+2)^2)=sqrt(7)$

$x + 2 = \pm \sqrt{7}$ $x+2=+-sqrt(7)$

$x = \pm \sqrt{7} - 2$ $x=+-sqrt(7)-2$

How do I complete the square if f(x) = 3x^2 + 12x - 9f(x)=3x2+12x−9f(x) = 3x^2 + 12x - 9?