Question

im struggling with this question can someone please help me?
Solve. x^2 + 6x + 4 = 0

Answer 1

Quadratic equations?
What method are you comfortable with? Quadratic formula?

Answer 2

i tried factoring but it didnt work out right

Answer 3

can you complete a square?

Answer 4

Because it can't be factored to the best of my knowledge :D

Answer 5

if ur equation is
ax^2+bx+c=0

Answer 6

i struggle with completing a square

Answer 7

Well, we can fix that :)

Answer 8

then u will get
x=(-b+-√(b^2-4ca))/2a

Answer 9

the concept behind completing a square amounts to realizing what a generic complete square looks like:\[(x+n)^2=x^2+2n~x+n^2\]
solving for n is how we complete the square

given x^2 + 6x + ______ , compare 2n with 6 and solve for n

Answer 10

so n will equal 3?

Answer 11

yes, notice that a complete square adds n^2 to it ... n=3, n^2 = 9

add and subtract 9 from the original set up since: (9-9) = 0 we are not changing the value of anything, just the way it looks

Answer 12

x^2+6x+9-9+4=0

(x^2+6x+9 ) -9+4=0

(x+3)^2 -5=0

Answer 13

add 5 to each side, sqrt, then subtract the 3

Answer 14

(x+3)^2 -5=0

(x+3)^2 = 5

x+3 = +- sqrt(5)

x = -3 +- sqrt(5)

Answer 15

the general formula for completing a square can be considered as:

2n = b

n = b/2

add and subtract n^2 to complete ... n^2 = (b/2)^2

Answer 16

ok i think i got my answer:
x=+-2sqrt5

Answer 17

you cant just combine -3 with sqrt(5) like that

Answer 18

x^2 +6x + 4 = 0

x^2 +6x +9 - 9 + 4 = 0

(x+3)^2 -5=0

(x+3)^2 = 5

x+3 = +- sqrt(5)

x = -3 +- sqrt(5)

Answer 19

i am rewriting it out to resolve it

Answer 20

To be general when it comes to solving quadratic equations of the form
\[\Large \color{red}ax^2 + \color{blue}bx + \color{green}c =0\]
We will use the method of 'completing the square'
First, we need to have the \(x^2\) alone, with no coefficient, so the first step is to divide everything by \(\color{red}a\)
\[\Large x^2 +\frac{\color{blue}b}{\color{red}a}x+\frac{\color{green}c}{\color{red}a}=0\]
Next, we subtract \(\Large \frac{\color{green}c}{\color{red}a}\) from both sides to that all the terms without \(x\) go on the right side...
\[\Large x^2 +\frac{\color{blue}b}{\color{red}a}x=-\frac{\color{green}c}{\color{red}a}\]


Now, to the actual completing of the square... recall that to complete the square given
\[\Large x^2 + \color{purple}px= k\]we take half of the coefficient of the lone \(x\) (the one with no exponent), in this case, \(\color{purple}p\), giving \(\Large \frac{\color{purple}p}{2}\) and then square it, yielding \(\Large \frac{\color{purple}p^2}{4}\) and add THAT to both sides of the equation...\[\Large x^2 +\color{purple}px+\frac{\color{purple}p^2}{4}=k+\frac{\color{purple}p^2}{4}\]such that the left side is now a perfect square...\[\Large \left(x+\frac{\color{purple}p}2\right)^2 = k+\frac{\color{purple}p^2}{4}\]


It so happens that in this case, our \(\color{purple}p\) value is \(\Large \frac{\color{blue}b}{\color{red}a}\). So half of that is \(\Large \frac{\color{blue}b}{2\color{red}a}\). Squaring this yields \(\LARGE \frac{\color{blue}b^2}{4\color{red}a^2}\) And this is now added to both sides of the equation \(\Large x^2 +\frac{\color{blue}b}{\color{red}a}x=-\frac{\color{green}c}{\color{red}a}\) giving us...

\[\Large x^2 +\frac{\color{blue}b}{\color{red}a}x+\frac{\color{blue}b^2}{4\color{red}a^2}=\frac{\color{blue}b^2}{4\color{red}a^2}-\frac{\color{green}c}{\color{red}a}\]

At this point, we may want to simplify the right-hand side of the equation into a single fraction, this can be done using their LCD, which happens to be \(\large 4\color{red}a^2 \)

\[\Large x^2 +\frac{\color{blue}b}{\color{red}a}x+\frac{\color{blue}b^2}{4\color{red}a^2}=\frac{\color{blue}b^2-4\color{red}a\color{green}c}{4\color{red}a^2}\]


Now, as intended, the left-hand side is now a perfect square, so...
\[\Large \left(x+\frac{\color{blue}b}{2\color{red}a}\right)^2=\frac{\color{blue}b^2-4\color{red}a\color{green}c}{4\color{red}a^2}\]

Taking the square root of both sides yields...\[\Large x+\frac{\color{blue}b}{2\color{red}a}=\pm\sqrt{\frac{\color{blue}b^2-4\color{red}a\color{green}c}{4\color{red}a^2}}\]
Simplifying...\[\Large x+\frac{\color{blue}b}{2\color{red}a}=\frac{\pm\sqrt{\color{blue}b^2-4\color{red}a\color{green}c}}{2\color{red}a}\]

Bringing the term \(\Large \frac{\color{blue}b}{2\color{red}a}\) to the right-hand side by subtracting from both sides gives
\[\Large x=-\frac{\color{blue}b}{2\color{red}a}\pm \frac{\sqrt{\color{blue}b^2-4\color{red}a\color{green}c}}{2\color{red}a}\]

And finally, combining, since they have the same denominator, we get...
\[\Large x= \frac{-\color{blue}b\pm \sqrt{\color{blue}b^2-4\color{red}a\color{green}c}}{2\color{red}a}\]


And this is the ever-notorious quadratic formula ^_^