How Can I Solve the Equation (5R^2 + 11)^(1/2) = -4R?

[adaptation]

Homework Statement

(5R²+11)^1/2=-4R

(It's actually written as a squareroot rather than being raised to the 1/2 power, but couldn't get latex to do it's thing.)

Homework Equations

None?

The Attempt at a Solution

I tried a couple different ways, none yeilding the correct answer. eg:
Squaring both sides to get
5R²+11=16R²

Combining like terms (16R²-5R²) to get
11=11R²

Without going any further, I can already tell that this is the wrong answer. Continuing, I'll get 1 rather than -1 (the solution when I solve graphically, also in the back of the book)

I also tried to begin by dividing both sides by -4. That didn't really get me anywhere.

Edit: I thought I should elaborate on that...
Dividing by -4
[(5R²+11)^1/2]/-4=R

Getting R by itself doesn't really change anything for me. I still need to "get rid" of the square root. It also gives me an unpleasant fraction to work with. I'd rather not go that route if I don't have to.

I've got some kind of mental block that won't let me see what I'm doing wrong. You don't have to give me all the steps to solve the problem, maybe just tell me what I did wrong and point me in the right direction.

Thanks in advance.

 

[AJKing]

\sqrt{5x^2+11}=-4x

5x^2+11=16x^2

11=11x^2

\sqrt{1}=x

\pm1=x

^ Are you sure that's not the answer in your book?

 

[AJKing]

\sqrt{5x^2+11}=-4x

[. . .]

\pm1=x

I missed a vital step.

I know what x equals, so now running the original equation with the values of x, I can see that x can only be -1.

If it's a positive one, then you're stating that you've taken a square root and returned a negative value.

Thus:

x = -1

 

[adaptation]

\sqrt{5x^2+11}=-4x

5x^2+11=16x^2

11=11x^2

\sqrt{1}=x

\pm1=x

^ Are you sure that's not the answer in your book?

Thanks AJKing! I made a stupid mistake. \sqrt{1} should be \pm1. My failure to solve this problem makes much more sense now that you've pointed this out to me. I forgot the negative solution...

I haven't run into a situation quite like this one before where only one of the solutions is a solution. (You guys can help me with the vocab there.) Is there a way to know (while solving the problem as opposed to checking) that only one answer is a solution?

 

[Mark44]

\sqrt{5x^2+11}=-4x

5x^2+11=16x^2

11=11x^2

\sqrt{1}=x

The line above should be \pm \sqrt{1} = x

\sqrt{1} = 1
The square root of 1 has one value, contrary to what many people mistakenly think.

\pm1=x

^ Are you sure that's not the answer in your book?

 

[adaptation]

The square root of 1 has one value, contrary to what many people mistakenly think.

Wait, what? I thought (-1)²=1. If \sqrt{1} doesn't have a negative solution, how can the solution to my original equation be negative?

I'm lost now.

 

[Mark44]

Wait, what? I thought (-1)²=1.

Yes, this is true.

If \sqrt{1} doesn't have a negative solution, how can the solution to my original equation be negative?

I'm lost now.

The equation x² = 1 has two solutions: one of them is positive (x = 1) and one of them is negative (x = -1).

If you wrote this:
x = \sqrt{1}
you are listing only one number; namely, 1.

Every positive, real number has two square roots - a positive square root, and a negative square root. The expression \sqrt{x}, where x >= 0, represents the principal (or positive) square root of the number inside the radical.