solve algebraically: x-3=radical x^2+3x?

x - 3 = √(x² + 3x)

(x - 3)² = x² + 3x

x² - 6x + 9 = x² + 3x

x² - x² - 6x - 3x + 9 = 0

-9x + 9 = 0

9x = 9

x = 1 <==================

If you think of math problems as snakes in the grass, then most of them are harmless garter snakes and black snakes. This one is a poisonous viper. 2 of the 10 answers preceding mine have successfully de-fanged the viper, namely:

Ed I and vahucel.

The other 8 are feverishly searching for the antidote.

The answer is there is no solution.

In addition to the explanations those 2 have given, I would only add that when you square (or take any even power of) both sides of an equation, you ALWAYS have to watch out for what happened here -- that you can introduce a solution that didn't work in the original equation.

Squaring loses some information, namely, that both sides had to have the same sign before you squared. In this case, the RHS was + or 0, so the LHS had to be as well, meaning that x had to be ≥ 3. In other words, the original equation requires that x be ≥ 3, because the RHS is a radical and can't be negative.

So when the answer comes back as x=1, right away you should recognize that this violates that requirement, so it has to be tossed out, leaving you with no solution.

EDIT:

Captain Matticus raises an interesting point.

It has been decades since I was in school, but I suspect this hasn't changed. Still, I suppose it's possible, because it involves a mathematical convention that could be taken in either of two ways without inconsistency.

So, Amee, I leave it to you, since it is your problem that you posed to us. If your math course takes the meaning of a radical to include both the + and - roots, then Captain Matticus is right, x=1 is a solution, and "sqrt" with this meaning is not a function. This raises other concerns whenever any radicals appear in expressions, so it seems unlikely to me.

If, on the other hand, "sqrt" IS taken as a function, and therefore can take only one value for each value of its argument, then ED and vahucel are right, and the equation has no solution.

This (definition of "sqrt" or "radical") is something that should have been touched upon in the course you're taking that generated this question; or if not, then it must have been assumed as having been covered in one of the prerequisite courses, before you started this course. It should have been introduced no later than Algebra I, more likely in a general math course preceding algebra.

It's your call, Amee. If you're uncertain, ask your teacher: "In this course, when we write, √100, does it mean +10 only, or does it mean both +10 and -10?"

Let us know what you find out.

I consider when you wrote radical... it is sqrt (square root)

So x- 3 = sqrt(x^2+3x).... put power 2 in both side.

(x-3)^2 = [sqrt(x^2 + 3x)]^2 ==> x^2 -6x + 9 = x^2 + 3x ==>

x^2 -6x + 9 -x^2 -3x = 0 ==> -9x + 9 = 0 ==> x = 1

When we put power in both side of the equation it can result in stranges roots, so we must test the root in the original equation. Put x=1 and get:

1-3 = sqrt(1^2 +3(1)) ==> -2 = sqrt(4) it is not true because sqrt(4) = 2 , only the positive number.

Then x=1 is not a root, then the equation has no root.

x-3 = √(x^2+3x)

x^2-6x+9 = x^2+3x

x^2 - 6x - 3x + 9 = x^2

x^2-9x+9 = x^2

x-3x+3 = x

1x-3x+3 = x

-2x+3 = x

3 = x+2x

3 = 3x

x = 1 <------Answer

Blessings

x = 1 seems to be the answer, UNTIL YOU TRY TO CHECK IT!

The solution does not check, so there is no solution.

-2 ≠ √4

The two sides, if graphed as equations, do not intersect.

Your question is somewhat ambiguous, however I assume you mean:

x-3=sqrt(x^2+3x)

So we first square both sides:

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

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

Then we expand the left side:

x^2-6x+9=x^2+3x

And subtract x^2:

-6x+9=3x

Collect like terms:

9=3x+6x

9=9x

And divide by 9:

x=1.

I don't see how this applies to trig, though...

I assume the radical is a square radical?

x-3 = radical(x^2+3x)

square both sides:

x^2 -6x+9 = x^2+3x

subtract x^2 from both sides:

-6x+9=3x

add 6x to both sides:

9=9x

divide both sides by 9

x=1

x - 3 = sqrt(x^2 + 3x) square both sides

x^2 - 6x + 9 = x^2 + 3x

9 = 9x

x = 1

Only the negative value of the sqrt(4) is a solution.

square both sides

(x - 3) = sqrt(x^2 + 3x)

Square both sides

x^2 - 6x + 9 = x^2 = 3x

9 = 9x

1 = x

EDIT:

Once again, Ed decided to do something that wasn't requested. There is no point in this problem where it is stated "Graph y = x - 3 and y = (x^2 + 3x)^(1/2) and find when they intersect." Had any variable (other than x or y) been used, would Ed have been so quick to graph the expressions and see if they intersect? I don't think so.

With that being said, all positive numbers have 2 roots, one being a positive root and the other being a negative root. Therefore, x = 1 is a perfectly acceptable answer in this instance, because:

1 - 3 = -2

sqrt(1 + 3) = sqrt(4) = +/- 2

-2 = -2

Q.E.F.D.

The F is for something amazing.

P.S. I'm only mentioning Ed because he has sent me e-mails in the past where he argues that I made a mistake in problems identical to this one (or at least the principles are identical). Hey, Ed! A graphing calculator is a relatively new invention. Newton didn't need one to find out that positive numbers can have negative roots!