# Completing the square problem

The problem i am supposed to solve for $x$ by completing the square:

$3x^2+9x+5 = 0$

step 1. move constant to right: $3x^2+9x\quad\quad+5 = 0$

step 2. divide by $3$: $x^2+3\quad\quad+\frac{5}{3}$

step 3. $(\frac{1}{2}b)^2$: $(\frac{1}{2}\cdot 3)^2 = \frac{9}{4}$

step 4. add and subtract result from step 3 into equation: $(x^2+3+\frac{9}{4}) -\frac{9}{4} + \frac{5}{3} = 0$

step 5. simplify: $(x+\frac{3}{2})^2 -\frac{7}{12} = 0$

step 6. subtract constant on both sides: $(x+\frac{3}{2})^2 = \frac{7}{12}$

step 7. solve for $x$: $x = -\frac{3}{2} + \sqrt{\frac{7}{12}}$ or $-\frac{3}{2} - \sqrt{\frac{7}{12}}$

Wolfram seems to be telling me my answer is wrong.

the correct answer listed is: $x = \frac{-9-\sqrt{21}}{6}$ or $x = \frac{\sqrt{21}-9}{6}$

If anyone could let me know where my error is I would greatly appreciate it!

• You dropped an $x$ in the passage from step 1 to step 2 (and in step 4, but this got corrected in step 5), but that didn't lead to a mistake. In fact up to that minor point, everything's right, just observe that $\sqrt{7/12} = \sqrt{21/36} = \sqrt{21}/6$. – t.b. Feb 9 '11 at 19:35
• hmm, how are they equal – Matt Feb 9 '11 at 19:42
• $\sqrt{\frac{7}{12}} = \sqrt{\frac{7 \cdot 3}{12\cdot 3}} = \sqrt{\frac{21}{36}} = \frac{\sqrt{21}}{\sqrt{36}} = \frac{\sqrt{21}}{6}$ – t.b. Feb 9 '11 at 19:49
• ah, rationalize the denominator – Matt Feb 9 '11 at 19:51

1. $3x^2+9x+5$
2. Factor out the leading coefficient from the variable terms: $3(x^2+3x)+5$
3. Determine the desired perfect-square constant term: $(\frac{1}{2}\cdot 3)^2=\frac{9}{4}$
4. Add and subtract the desired constant term inside the parentheses: $3(x^2+3x+\frac{9}{4}-\frac{9}{4})+5$
5. Factor the perfect square and distribute the leading coefficient: $3(x+\frac{3}{2})^2-3\cdot\frac{9}{4}+5$
6. Combine the constant terms: $3(x+\frac{3}{2})^2-\frac{7}{4}$