I have a math question.

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I have a math question.

What would the answer to this problem be: Solve the Equation: the square root of (2x+5) + the square root of (x) = 2

This open post was written 3 weeks, 4 days ago | V/U/S: 134, 25, 7 | Edit Post | Leave a reply | Report Post

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fractal.scatter 288 #
An Unknown Location | 3 weeks, 4 days ago (2 minutes after post)

sqrt(2x+5) + sqrt(x) = 2
2x+5+x=4
3x+5=4
You get the idea…

polkadotsocksxd 0 #
An Unknown Location | 3 weeks, 4 days ago (4 minutes after post)

That wouldn’t work though, because of FOIL, so it would really be somethign along the lines of
sqrt(2x+5) + sqrt(x) = 2
2x+5+(sqrt(2x+5)*sqrt(x))+(sqrt(2x+5)*sqrt(x))+x=4
but I sorta forget how to do that part and was wondering if there was an easier way…

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Arkia Valkair 8 #
An Unknown Location | 3 weeks, 4 days ago (10 minutes after post)

I don’t know of any easier ways. Basically it goes
sqrt(2x + 5) + sqrt(x) = 2
2x + 5 + 2 * sqrt(2x^2 + 5) + x = 2
3x + 3 + 2 * sqrt(2x^2 + 5) = 0
3x + 3 = -2 * sqrt(2x^2 + 5)
(3x + 3)^2 = (-2 * sqrt(2x^2 + 5)^2
…and so on…

dave1971199 1 #
An Unknown Location | 3 weeks, 4 days ago (16 minutes after post)

You look like you have a positive number and it really means 2×5. You normally would use +5 when you have a negative number.

I.E. (-2) x (+4)

I would suspect that (2x+5) is 10 and squared would be 100.

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fractal.scatter 288 #
An Unknown Location | 3 weeks, 4 days ago (16 minutes after post)

Righto. You’re right it wouldn’t work. But you have an idea of what you’re doing. Reduce the power 1/2 by taking logs.

brunorivera200 0 #
An Unknown Location | 3 weeks, 4 days ago (20 minutes after post)

The left side of this formula follows as:

2X+5+(2*sqrt(2x+5)*sqrt(x)+x=4

next, we combine the X coefficients:

3x+5+2*sqrt(2x+5)*sqrt(x)=4

now, we combine the constants:

3X+(5-4)+2*sqrt(2x+5)*sqrt(5)=0

3x+1+2*sqrt(2X+5)*sqrt(x)=0

There is still the difficulty of solving for the terms within sqare root operands!

fractal.scatter 288 #
An Unknown Location | 3 weeks, 4 days ago (25 minutes after post)

Well this is what I got;
(2x+5)^1/2 + x^1/2 = 2
1/2.ln(2x+5) + 1/2.ln(x) = ln(2)
1/2.ln(2x^2+5x) = ln(2)
ln(2x^2+5x) = 2.ln(2)
2x^2+5x = exp(4)
Rearrange to give
2x^2+5x-exp(4)=0
And solve using the quadratic formula to give

x = 5/4 +/- sqrt(25-16exp(2))/4

brunorivera200 0 #
An Unknown Location | 3 weeks, 4 days ago (25 minutes after post)

Arkia vaklar is on to something, by separating the square root expressions to one side of the equals sign and the rest of it to the other.

brunorivera200 0 #
An Unknown Location | 3 weeks, 4 days ago (34 minutes after post)

As I was saying…

-(3x+1)/2=sqrt(2x+5)*sqrt(x)

Then square both sides of the equation. Should I go on?

fractal.scatter 288 #
An Unknown Location | 3 weeks, 4 days ago (37 minutes after post)

brunorivera200 wrote:
As I was saying…

-(3x+1)/2=sqrt(2x+5)*sqrt(x)

Then square both sides of the equation. Should I go on?

Go on then. I’m intrigued.

Central 141 #
An Unknown Location | 3 weeks, 4 days ago (39 minutes after post)

fractal.scatter wrote:
Well this is what I got;
(2x+5)^1/2 + x^1/2 = 2
1/2.ln(2x+5) + 1/2.ln(x) = ln(2)
1/2.ln(2x^2+5x) = ln(2)
ln(2x^2+5x) = 2.ln(2)
2x^2+5x = exp(4)
Rearrange to give
2x^2+5x-exp(4)=0
And solve using the quadratic formula to give

x = 5/4 +/- sqrt(25-16exp(2))/4

DUH!

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fractal.scatter 288 #
An Unknown Location | 3 weeks, 4 days ago (41 minutes after post)

Central wrote:

fractal.scatter wrote:
Well this is what I got;
(2x+5)^1/2 + x^1/2 = 2
1/2.ln(2x+5) + 1/2.ln(x) = ln(2)
1/2.ln(2x^2+5x) = ln(2)
ln(2x^2+5x) = 2.ln(2)
2x^2+5x = exp(4)
Rearrange to give
2x^2+5x-exp(4)=0
And solve using the quadratic formula to give

x = 5/4 +/- sqrt(25-16exp(2))/4

DUH!

I’m not sure how I’m supposed to take that ‘DUH’?

Central 141 #
An Unknown Location | 3 weeks, 4 days ago (42 minutes after post)

fractal.scatter wrote:

Central wrote:

fractal.scatter wrote:
Well this is what I got;
(2x+5)^1/2 + x^1/2 = 2
1/2.ln(2x+5) + 1/2.ln(x) = ln(2)
1/2.ln(2x^2+5x) = ln(2)
ln(2x^2+5x) = 2.ln(2)
2x^2+5x = exp(4)
Rearrange to give
2x^2+5x-exp(4)=0
And solve using the quadratic formula to give

x = 5/4 +/- sqrt(25-16exp(2))/4

DUH!

I’m not sure how I’m supposed to take that ‘DUH’?

DUH as in, “DUH! That was soo easy even a rock could figure it out… thank you captain obvious!” said in a sarcastic tone…

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fractal.scatter 288 #
An Unknown Location | 3 weeks, 4 days ago (46 minutes after post)

Central wrote:
DUH as in, “DUH! That was soo easy even a rock could figure it out… thank you captain obvious!” said in a sarcastic tone…

Okay.

littlenick 126 #
An Undisclosed Location | 3 weeks, 4 days ago (47 minutes after post)

After reading this post and the replies that went with it, my brain exploded!!!!

Ugh!!!!

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swirly51 0 #
An Unknown Location | 3 weeks, 4 days ago (48 minutes after post)

X will be equal to -1/3.
First you square both sides to give you 2x + 5 + x = 4
3x = -1
x = -1/3

Anonymous #
3 weeks, 4 days ago (48 minutes after post)

Yeah mine too T-T
theres like 5 different answers here ….

fractal.scatter 288 #
An Unknown Location | 3 weeks, 4 days ago (50 minutes after post)

swirly51 wrote:
X will be equal to -1/3.
First you square both sides to give you 2x + 5 + x = 4
3x = -1
x = -1/3

Er… no.

Arkia Valkair 8 #
An Unknown Location | 3 weeks, 4 days ago (1 hour, 8 minutes after post)

swirly51 wrote:
X will be equal to -1/3.
First you square both sides to give you 2x + 5 + x = 4
3x = -1
x = -1/3

(a + b)^2 is not the same as a^2 + b^2. Things do not work that way.

swirly51 0 #
An Unknown Location | 3 weeks, 4 days ago (1 hour, 12 minutes after post)

my answer is correct. you go square all of both sides then solve for x. haha i had a 96% in advanced functions and 95% in calculus and vectors. i know what im doing.

swirly51 0 #
An Unknown Location | 3 weeks, 4 days ago (1 hour, 14 minutes after post)

ohhhhhhhh HAHA wow im an idiot. never mind im totallllllly wrong

Anonymous #
3 weeks, 4 days ago (1 hour, 15 minutes after post)

rofl… fail much? xD

fractal.scatter 288 #
An Unknown Location | 3 weeks, 4 days ago (1 hour, 15 minutes after post)

swirly51 wrote:
ohhhhhhhh HAHA wow im an idiot. never mind im totallllllly wrong

Yeah I hate iot when that happens lol.

Arkia Valkair 8 #
An Unknown Location | 3 weeks, 4 days ago (1 hour, 16 minutes after post)

swirly51 wrote:
ohhhhhhhh HAHA wow im an idiot. never mind im totallllllly wrong

Haha, it happens to all of us.

Central 141 #
An Unknown Location | 3 weeks, 4 days ago (1 hour, 27 minutes after post)

DUH!

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