#1

HELP!!!

#2

Try squaring both sides.

#3

sq root of x + sq root of 5= -2x +3?

#4

-1

#5

Am I the only one who can't see what pancakes needs help with?

#6

sq root of x + sq root of 5= -2x +3

No. If you add 2x to both sides it would be positive 2x + 3. Plus that doesn't really get you anywhere.

#7

Well, let's see.

Those snake things are called "numbers" and the two sticks crossed together are called "X", and that big pointy thing to the left is there so they don't get wet when it rains.

Those snake things are called "numbers" and the two sticks crossed together are called "X", and that big pointy thing to the left is there so they don't get wet when it rains.

#8

Halp!

#9

ok. its -4x^2 + x = 4

#10

add 2x to each side

square both sides..... x+5= 4x^2+12x+9

then combine like terms and factor

EDIT: i dont think it factors, so use the quadratic formula

square both sides..... x+5= 4x^2+12x+9

then combine like terms and factor

EDIT: i dont think it factors, so use the quadratic formula

*Last edited by AcsticRckr89 at Sep 3, 2007,*

#11

Its a quadratic if you square it, isn't it ? Make it equal to 0 if so .

#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?

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?

#13

problem is, i cant factor

4x^2 + 11x + 4 = 0

EDIT:

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

{i would have to solve for x.}

4x^2 + 11x + 4 = 0

EDIT:

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

{i would have to solve for x.}

#14

I refuse to help because you spelt it wrong.

#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

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.

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

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

#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

> 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

*Last edited by Kaze3 at Sep 3, 2007,*

#19

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.

Gracias Señor.

This is for you.

#20

.

*Last edited by Zackie EL at Mar 13, 2017,*

#21

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

#22

One free internet? omgz!

#23

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.

#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

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.

#26

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