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sq root of x + sq root of 5= -2x +3?
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ok. its -4x^2 + x = 4
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Its a quadratic if you square it, isn't it ? Make it equal to 0 if so .
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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

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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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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I refuse to help because you spelt it wrong.
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
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.
Gather everything to one side:

4x^2 - x + 4 = 0

(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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Pooping is well good though, to be fair.

I've got a handle on the fiction.

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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,
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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.
Last edited by Zackie EL at Mar 13, 2017,
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
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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.
Dear God, do you actually answer prayers?

Yes, but only in a way indistinguishable from random luck or the result of your own efforts.
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.

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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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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