Question

x^2-14x-4=0, solve by completing the square

Answer 1

hmmm, lets approach this from the perfect (complete) square concept

Answer 2

(x+a)^2 is a perfect square

(x+a)^2 = x^2 + 2a + a^2

given: x^2-14x-4

when does 2a = -14?

Answer 3

and therefore, what is a^2?

to complete the square, we need to add a useful form of zero to this. (a^2-a^2) = 0

Answer 4

ok i think i have figured this site out now.. these equations are blowing my mind

Answer 5

they do get easier after some practice :)

Answer 6

except for the fraction ones ... those are never easy to most people

Answer 7

lord i hope so..i was getting throught them fine and bam i'm like a deer in the headlgihts

Answer 8

so, following my setup, what do we get for "a" fromthis?

2a = -14

Answer 9

a=-7

Answer 10

good, "a" will always be half the "x" term of the poly given

what is a^2 knowing that a = -7?

Answer 11

ok so where did the 2 come in

Answer 12

is it from the 4???

Answer 13

a "complete square" is what we are trying to work out here. A complete square is a perfect square.

perfect squares are: 1^2, 2^2, 3^2 ... if using numbers

if using a binomial of the form (x+a) then we just square it: (x+a)^2 is a perfect square

Answer 14

(x+a)^2 = (x+a) (x+a) = x^2 +2a +a^2 ; if you wanna use "foil"

Answer 15

this gives a general setup to compare with the specific poly given:

x^2 -14x +a^2 -a^2 - 4
x^2 +2a +a^2 + _______

in order to "complete the square" we add and subtract a^2 to the given poly

Answer 16

since a is -7, lets add and subtract (-7)^2 = 49 from from it; why? (becasue we dont want to change the value of the poly, just the way it looks; so adding zero helps us out)

x^2 -14x +a^2 -a^2 - 4
x^2 -14x +49 - 49 - 4

group the perfect square

(x^2 -14x +49) - 49 - 4

and rewrite it in its (x+a)^2 form

(x-7)^2 - 53 = 0

Answer 17

ok so since x-7 in brackets it stays x-7 and i add 53 to both sides right

Answer 18

you can yes

Answer 19

ok whats the easiest and simple way to do this

Answer 20

well, completing the square is the proof to the quadratic formula

Answer 21

what is i make 53 a sq root and it would became 7sqrt53

Answer 22

*if

Answer 23

7 +/-sqrt53 right?

Answer 24

ugh i've confuesd myself

Answer 25

\[ax^2+bx+c = 0\]
\[x^2+\frac bax+\frac ca = 0\]
\[x^2+\frac bax+(\left(\frac b{2a}\right)^2-\left(\frac b{2a}\right)^2)+\frac ca = 0\]
\[\left(x^2+\frac bax+(\frac b{2a})^2\right)^2-\left(\frac b{2a}\right)^2+\frac ca = 0\]
\[\left(x+\frac b{2a}\right)^2-\left(\frac b{2a}\right)^2+\frac ca = 0\]
\[\left(x+\frac b{2a}\right)^2-\frac {b^2}{4a^2}+\frac ca = 0\]
\[\left(x+\frac b{2a}\right)^2-\frac {b^2}{4a^2}+\frac {4ac}{4a^2} = 0\]
\[\left(x+\frac b{2a}\right)^2-\frac {b^2}{4a^2}+\frac {4ac}{4a^2} = 0\]
\[\left(x+\frac b{2a}\right)^2=\frac {b^2}{4a^2}-\frac {4ac}{4a^2} = 0\]
\[\left(x+\frac b{2a}\right)^2=\frac {b^2-4ac}{4a^2}\]
\[x+\frac b{2a}=\pm\sqrt{\frac {b^2-4ac}{4a^2}}\]
\[x=-\frac b{2a}\pm\sqrt{\frac {b^2-4ac}{4a^2}}\]
\[x=-\frac b{2a}\pm\frac {\sqrt{b^2-4ac}}{2a}\]
\[x=-\frac {-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 26

that is correct: x = 7 +/-sqrt53 right

Answer 27

dont confuse the "a" in the formula with the random letter i choose to begin the post with :)

Answer 28

it kicked it out on my internet college math cours, it threw up up the \[\left(\begin{matrix}b \\ 2\end{matrix}\right)^2\]

Answer 29

ax^2 + bx + c = 0
x^2 -14x -4 = 0

\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\[x=\frac{-(-14)\pm\sqrt{(-14)^2-4(1)(-4)}}{2(1)}\]
\[x=\frac{14\pm\sqrt{196+16}}{2}\]
\[x=\frac{14\pm\sqrt{212}}{2}\]
\[x=\frac{14\pm\sqrt{4*53}}{2}\]
\[x=\frac{14\pm2\sqrt{53}}{2}\]
\[x=7\pm\sqrt{53}\]

Answer 30

ah it is wanting two numbers instead of a sqrt

Answer 31

hmm, yeah, i have never been to good with formating things the way a program wants it formated :)

Answer 32

the example shows x=11+/- 5 sqrt5

Answer 33

if you can take a screenshot of the problem, i could get a glimpse of what it is expecting an answer to look like :)

Answer 34

i tried to copy and paste it and it won't let me i just sent an email to my professior, i'm waiting on her reply..thank you for your help tho. i'll send message once i talk to her

Answer 35

ok, good luck ;)