Question

Solve the quadratic equation by completing the square.
3x^2 + 12x + 39 = 0
It says I need two solutions but I'm really lost.. :l

Answer 1

Quadratic formula,Do you know?

Answer 2

http://blogs.discovermagazine.com/loom/files/2008/07/quadratic.jpg
a:3
b:12
c:39

Can you do it now? @JennyMae18

Answer 3

$$\bf
3x^2 + 12x + 39 = 0\\
\text{group the "x" terms first}\\
(3x^2 + 12x) + 39 = 0\\
\text{common factor of them}\\
3(x^2+ 4x) + 39 =0\\
\text{what do we need inside the parentheses}\\
\text{ to get a perfect square trinomial?}\\
\large \color{blue}{3(x^2+ 4x+\square?) + 39 =0}\\
$$

Answer 4

When I do the problem I got-2 + sqrt(-13) , 2 - sqrt(-13)

Answer 5

Let me check. :D

Answer 6

Ok :)

Answer 7

How did you do it? Can you show? @JennyMae18

Answer 8

Sure.
3x^2+12x+39=0
I subtract 39 from both sides
3x^2+12x =-39
Then I divide by 3
x^2+4x = -13
4 goes into 'c'
X^2+4x+4 = -13
(x+2)^2 = -13
(x+2) = sqrt(-13)
x = -2 + sqrt(-13) , -2 - sqrt (-13)

Answer 9

Maybe you're doing this with a different formula...:3 @jdoe0001 i'm confused,help.

Answer 10

well, I understood it as completing the square to get the "x" values

Answer 11

$$\bf
3x^2 + 12x + 39 = 0\\
\text{group the "x" terms first}\\
(3x^2 + 12x) + 39 = 0\\
\text{common factor of them}\\
3(x^2+ 4x) + 39 =0\\
\text{what do we need inside the parentheses}\\
\text{ to get a perfect square trinomial?}\\
\large \color{blue}{3(x^2+ 4x+\square?) + 39 =0}\\
$$

Answer 12

so, the value we need to complete the square is \(2^2 = 4\)
since is inside the parentheses, the "3" multiplies it, so we're really adding 12
keep in mind, that all we're doing is borrowing from our good friend Mr Zero, 0,
so if we borrow 12, we have to give back 12
+12-12 = 0
so
$$\bf
3(x^2+ 4x+2^2) + 39 =0\\
\text{so above, we added } 3\times 2^2 = 12\\
\text{so we have to substract 12}\\
3(x+2)^2-12+39 = 0 \implies 3(x+2)^2+27=0\\
3(x+2)^2=-27\implies (x+2)^2=-\frac{27}{3} \implies \large (x+2)^2=-9
$$

Answer 13

\[x ^{2}+4x+4=-13+4=-9\]
\[\left( x+2 \right)^{2}=\left( 3\iota \right)^{2}\]
\[x+2=\pm3\iota, x=-2\pm3\iota \]
\[x=-2+3\iota \]
\[x=-2-3\iota \]