View Full Version : Pesky Math Problem

The37thElement

March 24th, 2016, 05:30 PM

I've been working on this one for an hour:

http://imgur.com/Bi26TUk

I have to solve for x. When I try to, I get 13 = 25; however Wolfram Alpha is telling me that x = -1 or -6. I need to find the intermediary steps. Help would be much appreciated!

Meron

March 24th, 2016, 05:54 PM

Edit: Realized my attempt was incorrect, I'll try a bit more and see what I can do.

The37thElement

March 24th, 2016, 06:00 PM

(^x+10) + (^3-x) = 5

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

(^x) + (^x) = -2

(2x) = -2

x = -1

That's the best I can do. I never took this kind of math problems before, but I tried. Hope you find it useful.

^ is the square root sign.

I appreciate the effort.

The thing about your process is that you removed the constants from the square roots (you would have to square both sides in order to move terms around).

Vlerchan

March 24th, 2016, 06:40 PM

The37thElement

(x + 10)^1/2 + (3 - x)^1/2 = 5

You raise the entire sum to the power of four. If raised to just two the x's cancel.

(x + 10)^2 + (3 - x)^2 = 625

x + 10(x + 10) + 3 - x(3 - x) = 625

x^2 + 10x + 10x + 100 + 9 - 3x - 3x + x^2 = 625

2x^2 + 14x + 109 = 625

2x^2 + 14x = 516

x^2 + 7x - 258 = 0

Introduce -B formula.

x = 12.939 or x = -19.939

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I'm quite sure that's correct. I'd double check yourself though because I sort of just dashed through that so I could post correct first.

I use ^ to refer to a raised power and ^1/2 refers to the square root.

Reise

March 24th, 2016, 07:07 PM

I can't see the image but according to:

sqrt(x+10) + sqrt(3-x) = 5

Vlerchan's wrong as -1 works. I'm not a hundred percents sure this can be expressed as a second degree polynomial though.

You can't raise to the power like that. For example 3 + 4 equals 7 but 9 + 16 does not equal 49.

Vlerchan

March 24th, 2016, 07:28 PM

Vlerchan's wrong as -1 works. I'm not a hundred percents sure this can be expressed as a second degree polynomial though.

(x + 10)^1/2 + (3 - x)^1/2 = 5

Let x = -1.

(-1 + 10)^1/2 + (3 - 1)^1/2 = 3 + 2^1/2 = 3 + 1.41421

4.41421 =! 5

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You're correct that I went wrong though.

The correct answer is -6 but I still need to work out how it is reached.

(-6 + 10)^1/2 + (3 - -6)^1/2 = 2 + 3 = 5

Reise

March 24th, 2016, 07:38 PM

Sorry bra. -1 works as well.

When you did (3-x)^1/2 you forgot that with -1 => x = -1

Therefore it makes 3-(-1)=4.

Square root of 4 is 2 and 2+3=5.

Perhaps this would help the OP.

sqrt(x+10) + sqrt(3-x) = 5

=> [sqrt(x+10) + sqrt(3-x)]^2 = 25

Use (a+b)^2 = a^2 + 2ab + b^2 and that would be ok.

Vlerchan

March 24th, 2016, 08:03 PM

Sorry bra. -1 works as well.

When you did (3-x)^1/2 you forgot that with -1 => x = -1

Therefore it makes 3-(-1)=4.

Square root of 4 is 2 and 2+3=5.

Well shit, you're right. Apologies. I'm clearly not on form tonight.

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