Which of the following is a solution of x2 + 4x + 6?
−2 + i times the square root of 2
2 + i times the square root of 2
−2 + three i times the square root of 2
−2 + three i times the square root of 2

- anonymous

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- anonymous

you have a choice
you can use the quadratic formula or you can complete the square
you pick

- anonymous

actually there is a third choice, namely to cheat

- anonymous

how do i do that, any of it ?

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## More answers

- anonymous

do you know the quadratic formula?

- anonymous

i need a refresher.

- anonymous

the solutions to
\[ax^2+bx+c=0\] are
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
in your case \(a=1,b=4,c=6\) but also in your case completing the square is easier

- anonymous

lets solve
\[x^2+4x+6=0\] by completing the square
ready?

- anonymous

sorry i was working it out xD

- anonymous

ok take your time

- anonymous

so far I got |dw:1438546685018:dw|

- anonymous

\[4^2-4\times 6=-8\]

- anonymous

\[x=\frac{-4\pm\sqrt{-8}}{2}\]

- anonymous

then
\[\sqrt{-8}=\sqrt{4\times 2\times (-1)}=\sqrt{4}\sqrt{2}\sqrt{-1}=2\sqrt{2}i\]

- anonymous

giving you
\[\frac{-4\pm2\sqrt{2}i}{2}\] divide each term by \(2\) and get
\[-2\pm\sqrt{2}i\]

- anonymous

it would have been easier to complete the square
there would have been no denominator

- anonymous

is that supposed to be the answer ?

- anonymous

lol yes it is "supposed to be" but in fact they wrote
\[-2\pm i\sqrt{2}\]

- anonymous

xD are you able to help me out with more ?

- anonymous

sure

- anonymous

Solve x2 + 4x + 1 = 0.
x equals the quantity of 4 plus or minus square root 12 all over 2
x equals negative 4 plus or minus square root 3
x equals negative 2 plus or minus square root 3
x equals the quantity of 4 plus or minus square root 13 all over 2

- anonymous

lets complete the square for this one, unless you want to use the formula again

- anonymous

in fact i will just write down the steps for completing the square without skipping any
so you can try one on your own

- anonymous

\[x^2+4x=1=0\] subtract 1 from both sides get
\[x^2+4x=-1\]

- anonymous

alright

- anonymous

then half of 4 is 2, and 2 squared is 4 go right to
\[(x+2)^2=-1+4\] or
\[(x+2)^2=3\]

- anonymous

take the square root of both sides, don't forget the \(\pm\) get
\[x+2=\pm\sqrt3\]

- anonymous

subtract 2 from both sides to solve for x, get
\[x=-2\pm\sqrt3\] done

- anonymous

looks like option C in your choices

- anonymous

xD I thought I wa supposed to do it lol

- anonymous

Solve 6 over x minus 4 equals 4 over x for x and determine if the solution is extraneous or not.
x = -8, extraneous
x = -8, non-extraneous
x = 8, extraneous
x = 8, non-extraneous
what about this one xD

- anonymous

this is not a quadratic i guess

- anonymous

\[\frac{x}{x-4}=\frac{4}{x}\] right?

- anonymous

no that is not right
\[\frac{6}{x-4}=\frac{4}{x}\] hows that?

- anonymous

the x in the numerator is a 6

- anonymous

as the math teacher say "cross multiply" and get
\[6x=4(x-4)\] which is a linear equation
can you solve that one?

- anonymous

yea let me try xD

- anonymous

i will wait

- anonymous

x=-8?

- anonymous

yes

- anonymous

would it be extraneous?

- anonymous

how do i tell if its extraneous or non-extraneous?

- anonymous

if it does not make the denominator zero, then it is not extraneous
or put another way, if it does make the denominator zero, then it is extraneous

- anonymous

in this case it does not, so it is not extraneous

- anonymous

oh now i get it

- anonymous

\[\frac{6}{x-4}=\frac{4}{x}\] sustitute \(x=-8\) get
\[\frac{6}{-8-4}=\frac{4}{-8}\] or
\[-2=-2\] a perfectly good solution

- anonymous

well actually
\[-\frac{1}{2}=-\frac{1}{2}\] but you get the idea

- anonymous

you up for some more xD? Or you tired out?

- anonymous

go ahead

- anonymous

What is the axis of symmetry for f(x) = 2x2 + 8x + 8?
x = −2
x = −4
x = 4
x = 2

- anonymous

oops
axis of symmetry of \(y=ax^2+bx+c\) is
\[-\frac{b}{2a}\]

- anonymous

in your case \(a=2,b=8\)

- anonymous

let me know when you get
\[x=-2\]

- anonymous

what is c?

- anonymous

nothing?

- anonymous

c in your case is 8 but it does not figure in to the axis of symmetry
it is just
\[x=-\frac{b}{2a}\]

- anonymous

so 2?

- anonymous

no there is a minus sign in the formula

- anonymous

oh so x=-2?

- anonymous

\[x=-\frac{8}{2\times 2}\\
x=-2\]

- anonymous

What is the standard form equation of the line shown below?
−3x + 2y = 5
3x − 2y = −5
y − 4 = three halves(x − 1)
y = three halvesx + five halves

##### 1 Attachment

- anonymous

you need the slope first
you know how to find it?

- anonymous

y=mx+b?

- anonymous

that is the "slope intercept" form
not standard form
but in any case you need \(m\) (the slope) first

- anonymous

what is the standard form again ?

- anonymous

standard form will be one of your first two choices
\[ax+by=c \]

- anonymous

im not understanding how i would use it

- anonymous

in fact we don't have to do any work at all for this one since only choice B is in standard form
just pick that one and move on

- anonymous

Simplify open parentheses x to the 2 fifths power close parentheses to the 5 sixths power.
x to the 37 over 30 power
x to the 13 over 30 power
x to the 10 elevenths power
x to the 1 third power

- anonymous

i really appreciate you helping me <3

- anonymous

\[\huge (b^m)^n=b^{m\times n}\] is what you need for this one, i.e.multiply the exponents

- anonymous

yw

- anonymous

so what would the b be?

- anonymous

multiply
\[\frac{2}{5}\times \frac{5}{6}\]

- anonymous

in this case \(b\) is \(2\)

- anonymous

do i cross multiply?

- anonymous

\[\huge (2^{\frac{2}{5}})^{\frac{5}{6}}\]
\[\huge 2^{\frac{2}{5}\times \frac{5}{6}}\]
hell no

- anonymous

multiply just means multiply

- anonymous

so\[\frac{ 10 }{ 30 }\]

- anonymous

\[\frac{2}{5}\times \frac{5}{6}\] we can cancel stuff too

- anonymous

the five

- anonymous

\[\frac{2}{\cancel{5}}\times \frac{\cancel{5}}{6}=\frac{2}{6}\] is a start

- anonymous

then you can cancel the 2 as well

- anonymous

or you could cancel here
\[\frac{1\cancel{0}}{3\cancel{0}}=\frac{1}{3}\]

- anonymous

either way you get \(\frac{1}{3}\) and also you do NOT cross multiply

- anonymous

ok gotcha cx makes more sense now

- anonymous

whew
final answer
\[\huge 2^{\frac{1}{3}}\]

- anonymous

What is the simplified form of the quantity 12 z-squared minus 7z minus 12 over the quantity 3 z-squared plus 2z minus 8 ?
4z plus 3 over z plus 2
4z minus 3 over z plus 2
4z plus 3 over z minus 2
4z minus 3 over z minus 2

- anonymous

lol

- anonymous

wait wouldnt it be x1/3

- anonymous

\[\frac{12z^2-7z-12}{3z^2+2z-8}\] factor and cancel

- anonymous

hold on let me go back and look

- anonymous

yeah sorry
\[\huge x^{\frac{1}{3}}\] the base was x, not 2

- anonymous

ahh gotcha

- anonymous

for the last one you have to factor and cancel
i suck at factoring, but i am good at cheating
http://www.wolframalpha.com/input/?i=\frac{12z^2-7z-12}{3z^2%2B2z-8}

- anonymous

i dont understand how to use that xD

- anonymous

this will work better
http://www.wolframalpha.com/input/?i=%2812z^2-7z-12%29%2F%283z^2%2B2z-8%29
apparently it is
\[\frac{4z+3}{z+2}\]

- anonymous

A?

- anonymous

click on the link to the second one and see how i wrote it in wolfram

- anonymous

yeah A

- anonymous

questions are all over the map
must be an on line course ...

- anonymous

What is the simplified form of the seventh root of x times the seventh root of x times the seventh root of x times the seventh root of x?
this one would be \[\sqrt[7]{4x}\] right?

- anonymous

no

- anonymous

yea it is XD

- anonymous

and oh i thought it was

- anonymous

one answer is
\[x^{\frac{4}{7}}\]

- anonymous

another is
\[\sqrt[7]{x^4}\]

- anonymous

the power is four, not the coefficient

- anonymous

\[x\times x\times x\times x=x^4\]not \(4x\)

- anonymous

4 times the seventh root of x
x to the four-sevenths power
the seventh root of 4x
x to the seven fourths power

- anonymous

thes are the answers

- anonymous

\[\huge \sqrt[7]{x^4}=x^{\frac{4}{7}}\]

- anonymous

make that a B

- anonymous

Which of the following represents the zeros of f(x) = 6x3 + 25x2 - 24x + 5?
-5, one third, one half
5, -one third, one half
5, one third, - one half
5, one third, one half

- anonymous

cheat like crazy for this one

- anonymous

http://www.wolframalpha.com/input/?i=+6x3+%2B+25x2+-+24x+%2B+5%3D0

- anonymous

\[\{-5,\frac{1}{2},\frac{1}{3}\}\]

- anonymous

so A?

- anonymous

otherwise you have to figure out that it factors as \[(3 x-1) (2 x-1) (x+5)\]

- anonymous

yeah A

- anonymous

Use the graph below for this question:
What is the average rate of change from x = −1 to x = −2?

##### 1 Attachment

- anonymous

\[5-7=-2\]

- anonymous

although the question is confusing since it says "from -1 to -2" which is backwards

- anonymous

maybe they want you to say 2, i am not clear on that
the question is ill posed

- anonymous

not uncommon for an on line class btw

- anonymous

sorry i was talking to my brother

- anonymous

i gotta run now
good luck with the rest
keep posting in a new thread, i am sure you will get lots of answers

- anonymous

thank you so much!!!

- anonymous

yw any time!

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