anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
you have a choice you can use the quadratic formula or you can complete the square you pick
anonymous
  • anonymous
actually there is a third choice, namely to cheat
anonymous
  • anonymous
how do i do that, any of it ?

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More answers

anonymous
  • anonymous
do you know the quadratic formula?
anonymous
  • anonymous
i need a refresher.
anonymous
  • 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
  • anonymous
lets solve \[x^2+4x+6=0\] by completing the square ready?
anonymous
  • anonymous
sorry i was working it out xD
anonymous
  • anonymous
ok take your time
anonymous
  • anonymous
so far I got |dw:1438546685018:dw|
anonymous
  • anonymous
\[4^2-4\times 6=-8\]
anonymous
  • anonymous
\[x=\frac{-4\pm\sqrt{-8}}{2}\]
anonymous
  • anonymous
then \[\sqrt{-8}=\sqrt{4\times 2\times (-1)}=\sqrt{4}\sqrt{2}\sqrt{-1}=2\sqrt{2}i\]
anonymous
  • anonymous
giving you \[\frac{-4\pm2\sqrt{2}i}{2}\] divide each term by \(2\) and get \[-2\pm\sqrt{2}i\]
anonymous
  • anonymous
it would have been easier to complete the square there would have been no denominator
anonymous
  • anonymous
is that supposed to be the answer ?
anonymous
  • anonymous
lol yes it is "supposed to be" but in fact they wrote \[-2\pm i\sqrt{2}\]
anonymous
  • anonymous
xD are you able to help me out with more ?
anonymous
  • anonymous
sure
anonymous
  • 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
  • anonymous
lets complete the square for this one, unless you want to use the formula again
anonymous
  • 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
  • anonymous
\[x^2+4x=1=0\] subtract 1 from both sides get \[x^2+4x=-1\]
anonymous
  • anonymous
alright
anonymous
  • 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
  • anonymous
take the square root of both sides, don't forget the \(\pm\) get \[x+2=\pm\sqrt3\]
anonymous
  • anonymous
subtract 2 from both sides to solve for x, get \[x=-2\pm\sqrt3\] done
anonymous
  • anonymous
looks like option C in your choices
anonymous
  • anonymous
xD I thought I wa supposed to do it lol
anonymous
  • 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
  • anonymous
this is not a quadratic i guess
anonymous
  • anonymous
\[\frac{x}{x-4}=\frac{4}{x}\] right?
anonymous
  • anonymous
no that is not right \[\frac{6}{x-4}=\frac{4}{x}\] hows that?
anonymous
  • anonymous
the x in the numerator is a 6
anonymous
  • 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
  • anonymous
yea let me try xD
anonymous
  • anonymous
i will wait
anonymous
  • anonymous
x=-8?
anonymous
  • anonymous
yes
anonymous
  • anonymous
would it be extraneous?
anonymous
  • anonymous
how do i tell if its extraneous or non-extraneous?
anonymous
  • 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
  • anonymous
in this case it does not, so it is not extraneous
anonymous
  • anonymous
oh now i get it
anonymous
  • 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
  • anonymous
well actually \[-\frac{1}{2}=-\frac{1}{2}\] but you get the idea
anonymous
  • anonymous
you up for some more xD? Or you tired out?
anonymous
  • anonymous
go ahead
anonymous
  • anonymous
What is the axis of symmetry for f(x) = 2x2 + 8x + 8? x = −2 x = −4 x = 4 x = 2
anonymous
  • anonymous
oops axis of symmetry of \(y=ax^2+bx+c\) is \[-\frac{b}{2a}\]
anonymous
  • anonymous
in your case \(a=2,b=8\)
anonymous
  • anonymous
let me know when you get \[x=-2\]
anonymous
  • anonymous
what is c?
anonymous
  • anonymous
nothing?
anonymous
  • 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
  • anonymous
so 2?
anonymous
  • anonymous
no there is a minus sign in the formula
anonymous
  • anonymous
oh so x=-2?
anonymous
  • anonymous
\[x=-\frac{8}{2\times 2}\\ x=-2\]
anonymous
  • 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
  • anonymous
you need the slope first you know how to find it?
anonymous
  • anonymous
y=mx+b?
anonymous
  • anonymous
that is the "slope intercept" form not standard form but in any case you need \(m\) (the slope) first
anonymous
  • anonymous
what is the standard form again ?
anonymous
  • anonymous
standard form will be one of your first two choices \[ax+by=c \]
anonymous
  • anonymous
im not understanding how i would use it
anonymous
  • 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
  • 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
  • anonymous
i really appreciate you helping me <3
anonymous
  • anonymous
\[\huge (b^m)^n=b^{m\times n}\] is what you need for this one, i.e.multiply the exponents
anonymous
  • anonymous
yw
anonymous
  • anonymous
so what would the b be?
anonymous
  • anonymous
multiply \[\frac{2}{5}\times \frac{5}{6}\]
anonymous
  • anonymous
in this case \(b\) is \(2\)
anonymous
  • anonymous
do i cross multiply?
anonymous
  • anonymous
\[\huge (2^{\frac{2}{5}})^{\frac{5}{6}}\] \[\huge 2^{\frac{2}{5}\times \frac{5}{6}}\] hell no
anonymous
  • anonymous
multiply just means multiply
anonymous
  • anonymous
so\[\frac{ 10 }{ 30 }\]
anonymous
  • anonymous
\[\frac{2}{5}\times \frac{5}{6}\] we can cancel stuff too
anonymous
  • anonymous
the five
anonymous
  • anonymous
\[\frac{2}{\cancel{5}}\times \frac{\cancel{5}}{6}=\frac{2}{6}\] is a start
anonymous
  • anonymous
then you can cancel the 2 as well
anonymous
  • anonymous
or you could cancel here \[\frac{1\cancel{0}}{3\cancel{0}}=\frac{1}{3}\]
anonymous
  • anonymous
either way you get \(\frac{1}{3}\) and also you do NOT cross multiply
anonymous
  • anonymous
ok gotcha cx makes more sense now
anonymous
  • anonymous
whew final answer \[\huge 2^{\frac{1}{3}}\]
anonymous
  • 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
  • anonymous
lol
anonymous
  • anonymous
wait wouldnt it be x1/3
anonymous
  • anonymous
\[\frac{12z^2-7z-12}{3z^2+2z-8}\] factor and cancel
anonymous
  • anonymous
hold on let me go back and look
anonymous
  • anonymous
yeah sorry \[\huge x^{\frac{1}{3}}\] the base was x, not 2
anonymous
  • anonymous
ahh gotcha
anonymous
  • 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
  • anonymous
i dont understand how to use that xD
anonymous
  • 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
  • anonymous
A?
anonymous
  • anonymous
click on the link to the second one and see how i wrote it in wolfram
anonymous
  • anonymous
yeah A
anonymous
  • anonymous
questions are all over the map must be an on line course ...
anonymous
  • 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
  • anonymous
no
anonymous
  • anonymous
yea it is XD
anonymous
  • anonymous
and oh i thought it was
anonymous
  • anonymous
one answer is \[x^{\frac{4}{7}}\]
anonymous
  • anonymous
another is \[\sqrt[7]{x^4}\]
anonymous
  • anonymous
the power is four, not the coefficient
anonymous
  • anonymous
\[x\times x\times x\times x=x^4\]not \(4x\)
anonymous
  • 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
  • anonymous
thes are the answers
anonymous
  • anonymous
\[\huge \sqrt[7]{x^4}=x^{\frac{4}{7}}\]
anonymous
  • anonymous
make that a B
anonymous
  • 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
  • anonymous
cheat like crazy for this one
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=+6x3+%2B+25x2+-+24x+%2B+5%3D0
anonymous
  • anonymous
\[\{-5,\frac{1}{2},\frac{1}{3}\}\]
anonymous
  • anonymous
so A?
anonymous
  • anonymous
otherwise you have to figure out that it factors as \[(3 x-1) (2 x-1) (x+5)\]
anonymous
  • anonymous
yeah A
anonymous
  • anonymous
Use the graph below for this question: What is the average rate of change from x = −1 to x = −2?
1 Attachment
anonymous
  • anonymous
\[5-7=-2\]
anonymous
  • anonymous
although the question is confusing since it says "from -1 to -2" which is backwards
anonymous
  • anonymous
maybe they want you to say 2, i am not clear on that the question is ill posed
anonymous
  • anonymous
not uncommon for an on line class btw
anonymous
  • anonymous
sorry i was talking to my brother
anonymous
  • 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
  • anonymous
thank you so much!!!
anonymous
  • anonymous
yw any time!

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