Question

complete the sq.
x^2 +9x -14=0

Answer 1

\[x^2+9x=14\]
\[(x+\frac{9}{2})^2=14+\frac{81}{4}\] etc

Answer 2

Once and for all, let's deal with this kind of problem :)
First thing you want to do is bring everything without an "x" to the right side... like so...
\[\large x^2 + 9x -14 \color{red}{+14}= 0 \color{red}{+14}\]
\[\large x^2 + 9x = \color{blue}{14}\]Catch me so far?

Answer 3

yup :)

Answer 4

Okay, to the left, you'll see "two kinds" of x's
One with a small "2" on top...
\[\large \color{green}{x^2}+9x = 14\]and the other, with no small number on top\[\large x^2 + 9\color{green}x = 14\]
Right?

Answer 5

yup

Answer 6

Okay, we're interested with the number that accompanies the x with no number on top.
This number, to be precise
\[\large x^2 +\boxed9x =14\]

Answer 7

So what I want you to do with it, is to take half of the number (which I boxed) and then square that.

Answer 8

Can you do that?

Answer 9

4.5^2=20.25

Answer 10

You prefer 4.5?
Okay, then... if you're sure...
Now, this number, we have to ADD IT TO BOTH SIDES OF THE EQUATION.
\[\large x^2 + 9x \color{red}{+20.25}= 14 \color{red}{+20.25}\]

Answer 11

then on the right side its 34.25

Answer 12

Yes. And on the left, you actually have...
\[\large x^2 + 9x + \color{red}{4.5^2} = \color{red}{34.25}\]

The point of this was to turn the left side into what's called a PERFECT SQUARE TRINOMIAL.
Well, now it is, so please express it as the square of some binomial... :)

Answer 13

how would i solve it from there?

Answer 14

Well, remember the number I boxed?
\[\large x^2 + \boxed9x + 4.5^2 = 34.25\]

Answer 15

Half of this was...?

Answer 16

you are doing a much much better job than i can but i have to butt in and say you really want to work with fractions and not decimals

Answer 17

Understood. @minimoo what we have done so far is not wrong, strictly speaking, but as @satellite73 said, we better be working with fractions...
Let's go back to this step.
\[\large x^2 +\boxed9x = 14\]
Once again, I want you to take half of 9, and then square it, but this time please do it using fractions and not decimals. Can you do that?

Answer 18

4 1/2 * 4 1/2= 20 1/4

Answer 19

You just love taking me out of my comfort zone, don't you? :P
Don't use mixed-number fractions, just use improper ones, (ones with just a numerator and denominator)

Much, much easier that way, trust me.

Answer 20

so 25/4

Answer 21

How do you get that? :)
Focus...
\[\Large x^2 + \boxed 9 x = 14\]

And if you're taking half of 9, just put a 2 under it :P
\[\Large \text{half of 9}= \frac92\]

Answer 22

Now, square this fraction...

Answer 23

81/4

Answer 24

Yes. So much like earlier, we add 81/4 to BOTH SIDES OF THE EQUATION.
\[\Large x^2 + 9x \color{red}{+ \frac{81}4}=14 \color{red}{+\frac{81}4}\]

Answer 25

Can you simplify the right side before we proceed?

Answer 26

mhmm 137/4

Answer 27

You're fast o.O

Answer 28

haha simple math is easy

Answer 29

LOL

\[\Large x^2 + 9x +\frac{81}4=\color{red}{\frac{137}4}\]

Answer 30

So now, the left is what you'd call a perfect square trinomial.
Well... in general...
\[\Large \color{blue}{x^2 + bx + \left(\frac{b}2\right)^2=\left(x+\frac{b}2\right)^2}\]

Answer 31

So, can you express the left side as the square of some binomial? :)

Answer 32

x(x+9)?

Answer 33

NO, see, you have an 81/4 there, and that's not such a bad thing (we put it there, in the first place).
Maybe you'll see the light when you recall that.\[\Large \frac{81}4=\left(\frac92\right)^2\]

Answer 34

but then how would i factor it out?

Answer 35

\[\Large x^2 + 9x +\left(\frac92\right)^2=\frac{137}4\]
\[\Large \color{blue}{x^2 + bx + \left(\frac{b}2\right)^2=\left(x+\frac{b}2\right)^2}\]

Seeing the pattern?

Answer 36

yeah

Answer 37

So, the left side of the equation can be expressed as...?

Answer 38

a trinomial?

Answer 39

It is a trinomial, but you should express it as the square of some binomial... that was the whole point of adding 81/4 to both sides...

Answer 40

ugh x/ my

Answer 41

i don't see how i find the solut

Answer 42

solution that way

Answer 43

Okay, relax... take a deep breath...

Answer 44

And visualise this...
\[\Large \color{blue}{x^2 + \color{red}bx + \left(\frac{\color{red}b}2\right)^2=\left(x+\frac{\color{red}b}2\right)^2}\]
Are you feeling it...? :D

Answer 45

x(2x+3b/2)^2=(x+b/2)^2

Answer 46

No, relax... and just focus on this....
\[\Large \color{black}{x^2 + \color{red}bx + \left(\frac{\color{red}b}2\right)^2=\left(x+\frac{\color{red}b}2\right)^2}\]

Have you taken it in yet?

Answer 47

yes

Answer 48

Now, what if I were to replace b with a number, say, I don't know, 9?
\[\Large \color{black}{x^2 + \color{red}9x + \left(\frac{\color{red}9}2\right)^2=}\huge \color{blue}?\]

Answer 49

ok

Answer 50

Well, what happens to the right side?
(HINT: All we did was replace b with 9...)

Answer 51

sq root

Answer 52

NO, I told you to visualise this...
\[\Large \color{black}{x^2 + \color{red}bx + \left(\frac{\color{red}b}2\right)^2=\left(x+\frac{\color{red}b}2\right)^2}\]
So here
\[\Large \color{black}{x^2 + \color{red}9x + \left(\frac{\color{red}9}2\right)^2=\color{green}{\huge ?}}\]
What becomes of the right side if I just **replace b with 9** ?

Answer 53

x+9/2)^2

Answer 54

Much better.
\[\Large \color{black}{x^2 + \color{red}9x + \left(\frac{\color{red}9}2\right)^2=\left(x+\frac{\color{red}9}2\right)^2}\]

So here... this bit...
\[\Large \color{red}{x^2 + 9x +\left(\frac92\right)^2}=\frac{137}4\]

It may be replaced as follows
\[\Large \color{blue}{\left(x+\frac92\right)^2}=\frac{137}4\]

Answer 55

You following me?

Answer 56

yeah :)

Answer 57

Are you sure?

Answer 58

Okay, so at this point, not much left to do, get the square root of both sides of the equation... Can you do that?

Answer 59

yup :) x+3x+9/2=Sq rt 137/2

Answer 60

Where'd the 3x come from?
\[\Large \sqrt{\color{}{\left(x+\frac92\right)^2}}=\sqrt{\frac{137}4}\]

Answer 61

oh so just\[\sqrt{137}\]/2

Answer 62

Yeah, but the left side, where did you get that 3x? :/

Answer 63

sq rt of 9

Answer 64

\[\Large \sqrt{\color{}{\left(x+\frac92\right)^2}}={\frac{\sqrt{137}}2}\]

Answer 65

No, see, the square root sign, and the square (the little 2) just cancel out...
\[\Large\color{red}{\sqrt{\color{black}{\left(x+\frac92\right)^{\color{green}2}}}}={\frac{\sqrt{137}}2}\]

Answer 66

So you can remove them both...
\[\Large{\color{}{x+\frac92}}={\frac{\pm\sqrt{137}}2}\]

Answer 67

ok

Answer 68

And now, (I put the +- sign on the right, that's necessary, okay? because a square root may be positive or negative)

All that's left is bringing the 9/2 to the other side, and simplifying
\[\Large\color{}{x+\frac92\color{red}{-\frac92}}=\color{red}{-\frac92}+{\frac{\pm\sqrt{137}}2}\]
\[\Large{\color{}{x}}={\frac{\color{blue}{-9}\pm\sqrt{137}}2}\]

Answer 69

1.35 and -10.35?

Answer 70

Simplify it as you will but I usually stop at that point. :)

Answer 71

haha thanks so much :))

Answer 72

No problem :)