Best to do your own homework. The rules of this site are clear about homework problems. I can only suggest you expand the equation. It easily simplifies to an obvious answer. Good luck.

It's not my homework. I was just trying to help explain it to my little sis and to my dismay, i couldn't! Cos this very much looks like a simple question! But i had a breakthrough yesternight though...

Extra credit: find a second negative value of x that satisfies the equation.

Not a derivation, but how about 6, 2, -2 and -6 as four solutions? How did Tornado know they would turn out to be an arithmetic progression?

I didn't know ahead of time, only noticing it after the fact; it's just blind luck, and a "hint" that allows confirmation of the solution without giving much away. They can be expressed succinctly as ±4±2.

Further extra credit: find a second positive value of x to satisfy the equation.

Hint: since you want a negative x, 2|x| is equal to -2x. Both answers are nice round numbers.

Try to think: if x<0 there is always a number y>0 such that abs(x)=abs(y).

Thinking this way you find yourself only in the >0 domain and it becomes a lot easier to handle the equation.

To find the solution will be your problem. I only gave you the hint with the hope you will understand and use it the right way.

(-x)² = x²; i.e., positive real numbers have 2 square roots.

All solutions to the original equation are in arithmetic progression.

(x²)^0.5 = ±x; giving two more solutions, each using the negative square root.

Perhaps I'm missing something, but when I put x = 6 or -6 into the formula, I get 12, not 0.

Cheers........Codey

Only possible solutions are +/-2

When you take the negative square root (0.5 power), you get the other half of the graph.

From the Mathcad help section...

Square root

Keystroke: \

Returns the positive square root of x.

Tornado has already pointed it out. There are two answers to a square root, one positive (the answer that your calculator and software such as excel and mathcad gives you) and one negative.

Yes, I stand corrected. Forgot Mathcad only did positive square root. Here is a plot of 4 equations showing all 4 possibilities. Hope the color is clear enough to see the differences.

Apparently negative square roots are deemed off-topic, at least by one idiot.

ok for those of you who aren't grasping the negative root concept let's do a quick review...

x = y²

Imagine going up and down the Y-axis squaring each number to obtain the X value. Note there is nothing preventing you from squaring a negative Y value. The graph is a sideways parabola. Now take note that for each value of X there are TWO Y values (one positive and one negative).

Now let's rearrange our equation taking the square root of each side to put it into a standard form.

y = x^1/2

The graph doesn't change (it is still the same function). There are still two Y values for each and every X. Excel, Mathcad, and calculators are programmed to return only the POSITIVE root.

As Tornado has tried to point out, there are four equations here and hence four solutions...

(±x ± 2x = 6)

x + 2x = 6 (x = 2)

x - 2x = 6 (x = -6)

-x + 2x = 6 (x = 6)

-x - 2x = 6 (x = -2)

Wrong again. No square roots are being taken of negative numbers in this exercise; this is about the negative square root of positive numbers. In particular, √36 = +6 and also -6.

"A real number squared, then square-rooted is an alternate definition of the Absolute value function."

Where did that definition come from? It can't be correct. It doesn't work for any negative number.

I urge you to graph y² = x as I mentioned above. It won't take but a few minutes. Start at y = -10 square it and plot the value (100,-10). Add one to Y and continue until you get to y = 10. Then connect the dots.

You will see there are 2 values of Y for each and every X. (and note that all X values are non-negative).

Now take the square root of the function... so...

y = sqrt (x)

It is the same function. Nothing has changed. There are still 2 values of Y for each X.

Actually simplify the whole thing even more... let's read in words what this says...

x = sqrt (36)

It says, what number when multiplied by itself is equal to 36? both 6 and -6 satisfy this.

Expanding [(x+1)² - (2x+1)]^0.5 + 2|x| - 6 =0,

we get [x² + 2x +1 - 2x - 1]^0.5 + 2|x| - 6 =0

Or [x²]^0.5 + 2|x| = 6

Or ±x + 2|x| = 6

So either 3x = 6 (x = 2) or x = 6 (2 and 6 are the positive roots)

But if x = -2 or x = -6 the result is the same so -2 and -6 are the negative roots.

I'm glad you get the positive/negative root issue. There's only one more small step to make...

To "square root" a number is the same as taking it to the one-half power or in decimal form 0.5 (which is part of the original function).

sqrt(x) = x^0.5 = x^1/2

Good to go? If not, don't sweat it... I'll keep explaining until you get it.

I am quite taken aback how difficult are such equations for CR4 participants and how proud they are to suggest a solution. This is a sign. The last comment is not correct since it is NOT necessary to put a sign there is a definition by the equation and this should be enough. It was generally agreed upon NOT to give solutions but only to help by hints. The problem has been considered so complex that many looked at it as chalenge and wanted to show that they can solve it.

Take the initial equation: [(x+1)²-(2x+1)]^0.5 + 2|x| - 6 = 0

Substitute 2 for x and see if the left hand side equals 0.

[(2+1)²-(2*2+1)]^0.5 + 2|2| - 6 = [9-5]^0.5 + 4 - 6 = [4]^0.5 + 4 - 6,

but [4]^0.5 has two values +2 and -2.

So 2 + 4 - 6 = 0 and -2 + 4 - 6 = 4 thus 2 is valid solution.

Substitute -2 for x and see if the left hand side equals 0.

[(-2+1)²-(2*-2+1)]^0.5 + 2|-2| - 6 = [1+3]^0.5 + 4 - 6 = [4]^0.5 + 4 - 6,

but [4]^0.5 has two values +2 and -2.

So 2 + 4 - 6 = 0 and -2 + 4 - 6 = 4 thus -2 is valid solution.

Substitute 6 for x and see if the left hand side equals 0.

[(6+1)²-(2*6+1)]^0.5 + 2|6| - 6 = [49-13]^0.5 + 12 - 6 = [36]^0.5 + 12 - 6,

but [36]^0.5 has two values +6 and -6.

So 6 + 12 - 6 = 12 and -6 + 12 - 6 = 0 thus 6 is valid solution.

Substitute -6 for x and see if the left hand side equals 0.

[(-6+1)²-(2*-6+1)]^0.5 + 2|-6| - 6 = [25+11]^0.5 + 12 - 6 = [36]^0.5 + 12 - 6,

but [36]^0.5 has two values +6 and -6.

So 6 + 12 - 6 = 12 and -6 + 12 - 6 = 0 thus -6 is valid solution.

For those who are still unconvinced, another approach is as follows:

[(x+1)² - (2x[(x+1)² - (2x+1)]^0.5 + 2|x| - 6 =0 which is:

[(x)²]^0.5 = - 2|x| + 6 = 2x + 6 (since x is negative)

Squaring both sides,

x² = 4x² + 24x + 36

Or 3x² + 24x + 36 = 0

Or x² + 8x + 12 = 0

This can be factored into (x+2)(x+6) = 0, leaving us with the two negative roots (-2, -6). If positive roots are wanted, they would be found to be (2, 6) by a similar procedure.

This whole thread has bogged down on the question of negative square roots of positive numbers. Every positive number does indeed have two square roots. Some conventions define √x (or x^0.5) to be the positive square root only. Adhering to that would eliminate the ±6 solutions to the original problem. A "mathematically pure" view would still allow them, but this notational convention would rule them out. Therefore I relax the earlier insistence on including them.

For a negative number x, the two square roots are "±i√x", which is in quotes because + and - don't have a clear meaning in the complex plane. Hence my preference for keeping both square roots considered as equally legitimate.

Hello Tornado

Not sure about the last bit. I know what you mean, but strictly speaking it should be ±i√mod(x). If x is negative ±i√x = ±i*i*√mod(x) = ±√mod(x).

Also there's no difficulty plotting ±i√mod(x) in the complex plane. It gives 2 points on the imaginary axis, at distances ±√mod(x) from the origin.

Cheers........Codey

Sort of correct; I should have said ±i√|x|, or ±i√(-x). But mod(x) is another concept. There is no problem graphing the roots, but there is a problem in designating what is + or - in the complex plane. For instance, is 1-i positive or negative?

OK, by mod(x) I meant same as |x|. Like the OP, I didn't think I could type |x|, but I tried it and it worked!

The question "is 1-i positive or negative?" has no meaning in the complex domain. Also > and < have no meaning.

Codey