How do you complete the square for $3 x^{2} + 18 x + 5$ $3x^2 +18x + 5$ ?

Question

7637 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-05T08:49:03

$3 {(x + 3)}^{2} - 22$ $3(x+3)^2-22$

$3 x^{2} + 18 x + 5$ $3x^2+18x+5$

take out common factor to make coefficient $x^{2} = 1$ $x^2=1$

$= 3 (x^{2} + 6 x) + 5$ $=3(x^2+6x)+5$

now CTS as normal inside the bracket as follows

$= 3 (x^{2} + 6 x + 3^{2} - 3^{2}) + 5$ $=3(x^2+6x+3^2-3^2)+5$

$= 3 ({(x + 3)}^{2} - 9) + 5$ $=3((x+3)^2-9)+5$

$= 3 {(x + 3)}^{2} - 27 + 5$ $=3(x+3)^2-27+5$

$= 3 {(x + 3)}^{2} - 22$ $=3(x+3)^2-22$

Answer 2 · 2018-04-05T08:55:05

$3 {(x + 3)}^{2} - 22$ $3(x + 3)^2 - 22$

Before completing the square a must equal 1, so first we divide all terms by 3.

= $\frac{3 x^{2}}{3} + \frac{18 x}{3} + \frac{5}{3}$ $(3x^2)/3 + (18x)/3 + (5)/3$

= $3 (x^{2} + 6 x + \frac{5}{3})$ $3(x^2 + 6x + 5/3)$

= $3 (x^{2} + 6 x + {(\frac{6}{2})}^{2} - {(\frac{6}{2})}^{2} + \frac{5}{3})$ $3(x^2 + 6x +(6/2)^2 - (6/2)^2 + 5/3)$

= $3 ((x^{2} + 6 x + 9) - 9 + \frac{5}{3})$ $3((x^2 + 6x + 9) - 9 + 5/3)$

= $3 ((x^{2} + 6 x + 9) - \frac{27}{3} + \frac{5}{3})$ $3((x^2 + 6x + 9) - 27/3 + 5/3)$

= $3 ((x^{2} + 6 x + 9) - \frac{22}{3})$ $3((x^2 + 6x + 9) - 22/3)$

$(x^{2} + 2 x y + y^{2}) = {(x + y)}^{2}$ $(x^2 + 2xy + y^2) = (x + y)^2$ , so...

= $3 ({(x + 3)}^{2} - \frac{22}{3})$ $3((x + 3)^2 - 22/3)$

= $3 {(x + 3)}^{2} - 22$ $3(x + 3)^2 - 22$

How do you complete the square for 3x^2 +18x + 53x2+18x+53x^2 +18x + 5?