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Find the x-intercepts: 2(x-5)^2=17

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x= [17/2]^1/2 +5|dw:1345790422320:dw|
x-intercepts? this is an equation in one variable.... how do you do that?

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Other answers:

Am I suppose to distribute the 2?
ahh.... my bad.... because this is an equation in one variable, it is just a vertical line.... so solving for x will give you the only x-intercept....
since this is a quadratic function, you will get 2 values for the x-intercepts. read pp 126-129 in the link I've posted above.
2 x-intercepts? \[2(x-5)^2=17\]Divide both sides by 2\[(x-5)^2=\frac{17}{2}\]Take square roots for both sides \[(x-5)=\pm \frac{17}{2}\]Add 5 to both sides to get the answer.
I'm always late....
oh okay thank you that makes so much sense :)
hold on folks... this is an equation in x only.... this is a vertical line.... and not a function.... "Find the x-intercepts: 2(x-5)^2=17"
Wow. That's complicated. I thought all I had to do was to solve it but seems not :(
this is a function: y = 2(x-5)^2 this is not a function: 17 = 2(x-5)^2
parabolas always have 2 x-intercepts unless k=0, is it not?
But how to get the x-intercept(s)??? There is no y...
in any function, to find the x-intercepts, set y=0 and solve.... how are you gonna set y=0 in the equation 2(x-5)^2 =17 ????
Think the x-intercept is 8.5,0
Perhaps this would be the case? y = 2(x-5)^2-17 Put y=0 2(x-5)^2-17 = 0 2(x-5)^2=17 . . . Solve x to find the x-intercepts. Hmm...
dpalnc solved it, x=5 - sqrt(17/2) x=5+ sqrt(17/2)
compare these two equations: y = x + 3 and 17 = x + 3 that first one is a line and you can find the x-intercept by setting y=0 then solving 0 = x + 3. but that second equation is just a vertical line....
Eh?! Then, for 17=x+3, the x-intercept is 17-3 = 14?!
you have been trolled
dang this chrome browser.... x = 14 is a vertical line and that's where it crosses the x-axis.... but back to the problem... did i say 1 x-intercept? i mean two as panlac says.... :)
a vertical line that curves...obviously troll ^^
how 'bout two vertical lines... it's implied when u solve a quadratic you have to consider the positive and negative square root....
yeah buwahhahahahaha
ang troll ^^^
this goes against the teachings of the monks... the forefathers defined x-intercept as "the value of y when x is 0" however, there is no y....
this is not a function
right... so when you solve for x in that equation, you get two vertical lines... an equation in only 1 variable x is not a function...
no. this is just not a function. nothing more; nothing less
it's a relation, but not a function
x-intercepts occur in function only
when did i say it is a function?
I said it was, my mistake
you were saying it had an x-intercept
two values for x...
it has values for x...but no x-intercept
but the vertical line intercepts the x-axis...
yes. LG you are right "The point at which a curve or FUNCTION crosses the x-axis (i.e., when in two dimensions)."
like i said. you got trolled
cross or touches...
so are you saying that a circle (which is a curve) does not have x-intercepts?
yes it doesnt. it's not a function
circle is not a function
neither do ellipses
function = 1 to 1 value
so why can't you call the where a vertical line crosses the the x-axis the x-intercept?
any closed figure is not a function. it is a plane
this is going beyond the problem now... bottom line: x has two values
i agree to agree... if that makes sense....
hey LG... i thnk u scared off the asker.... or he/she got bored...
did she ask the problem to be graphed?
lol jk. let us just leave it
i would like to quote myself "you got trolled" relations have x-intercepts too
yes.... let's graph it....
this is really going beyond now... why can't we agree that the intercepts are where either the vertical line or horizontal line are touched or crossed? did I start the fire when I said it was a function? I retracted it so OS can be a better place for students once again...
sorry man... i just miss talkin to LG....
u know u'd make a great ambassador for keeping the peace....:)
so peace to everyone.....:)
I'd be damned...

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