#1
HELP!!!


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#3
sq root of x + sq root of 5= -2x +3?
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#4
-1
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#5
Am I the only one who can't see what pancakes needs help with?
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#7
Well, let's see.

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#9
ok. its -4x^2 + x = 4
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#11
Its a quadratic if you square it, isn't it ? Make it equal to 0 if so .
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#12
Add 2x to each side.

sqrt (x-5) = 2x+3

square each side to get x out of a radical

x-5 = (2x+3)(2x+3)

use FOIL method or squaring method, whatever

x-5 = 4x^2 + 6x + 6x + 9

x - 5 = 4x^2 + 12x + 19

Add 5 to each side

x = 4x^2 + 12x + 24


Do you have to solve for X, or do you need the simplest form of the equation?
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#13
problem is, i cant factor

4x^2 + 11x + 4 = 0

EDIT:

OH **** MABE IT ISNT FACTORABLE???

{i would have to solve for x.}

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#15
can be rewritten as (x+5)^0.5 -2x =3

i know that probably doesnt help much on this question but it will when you do calculus :P
#16
sq rt(x+5) - 2x = 3
sq rt(x+5)= 2x+3
(sq rt(x+5)^2)=(2x+3)^2
x+5=(2x+3)(2x+3)
x+5=4x^2+12x+9
0=4x^2+11x+4
4x^2+11x+4=0

Then you factor that, but it's unfactorable, so it's unsolvable. That's the answer.
#17
Gather everything to one side:

4x^2 - x + 4 = 0

Use the standard quadratic equation:

http://en.wikipedia.org/wiki/Quadratic_formula#Quadratic_formula
(couldn't be bothered to type it, it'll look a mess)

and then solve for the two values of x
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#18
sqrt(x+5) - 2x = 3

> sqrt(x+5) = 3 + 2x

> x + 5 = (3+2x)^2

> x + 5 = 9 + 4(x^2) + 12x

> 4(x^2) + 11x + 4 = 0

solve using quadtratic formula x = -b +/- sqrt(b^2 - 4ac) / 2a where a=4, b=11, c=4

gives crappy roots tho

ithink thats right neway
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Last edited by Kaze3 at Sep 3, 2007,
#19
Quote by Mzyz
sq rt(x+5) - 2x = 3
sq rt(x+5)= 2x+3
(sq rt(x+5)^2)=(2x+3)^2
x+5=(2x+3)(2x+3)
x+5=4x^2+12x+9
0=4x^2+11x+4
4x^2+11x+4=0

Then you factor that, but it's unfactorable, so it's unsolvable. That's the answer.



cool.

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#21
Quote by Mzyz
sq rt(x+5) - 2x = 3
sq rt(x+5)= 2x+3
(sq rt(x+5)^2)=(2x+3)^2
x+5=(2x+3)(2x+3)
x+5=4x^2+12x+9
0=4x^2+11x+4
4x^2+11x+4=0

Then you factor that, but it's unfactorable, so it's unsolvable. That's the answer.

Um...no. It just means you use the quadratic formula.

-11 +/- sqrt(11^2 - 4*4*4)
all divided by 8

Something like -.431 and -2.319
#23
Quote by Mzyz

Then you factor that, but it's unfactorable, so it's unsolvable. That's the answer.





People. You've got the whole damn Internet at your disposal.

If you couldn't even think of typing "solving quadratic equations" into Google and hitting the first (or first few) links, I don't give this generation much hope.
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Yes, but only in a way indistinguishable from random luck or the result of your own efforts.
#24
I just meant it's unfactorable for integers, which I'm sure is the case as I'm assuming the thread poster is in some sort of Calc or Precalc course in high school or early college.
#25
Quote by Mad Marius


People. You've got the whole damn Internet at your disposal.

If you couldn't even think of typing "solving quadratic equations" into Google and hitting the first (or first few) links, I don't give this generation much hope.


Thats the spirit, lower your expectations.

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#26
Quote by Mzyz
I just meant it's unfactorable for integers, which I'm sure is the case as I'm assuming the thread poster is in some sort of Calc or Precalc course in high school or early college.


Yeah, i totally forgot about imaginary numbers.

=P

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