x² + root(x²+3x+5) = 7 - 3x

Thanks :)

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- Feb 22nd 2007, 11:36 AMTrenttRadicals Problem
x² + root(x²+3x+5) = 7 - 3x

Thanks :) - Feb 22nd 2007, 12:24 PMJhevon
Ok, so this problem gets really messy--too messy to type some parts of it. If anyone out there has a shorter way of doing, be my guest and post it, cause this is the only way I see how. Here goes:

x^2 + sqrt(x^2 + 3x + 5) = 7 - 3x

=> sqrt(x^2 + 3x + 5) = -x^2 - 3x + 7

=> x^2 + 3x + 5 = (-x^2 - 3x + 7)^2.......................squared both sides

Here comes the monstrosity:

=> x^2 + 3x + 5 = x^4 + 3x - 7x^2 + 9x^2 + 3x^3 - 21x - 7x^2 - 21x + 49

Whew! Let's simplify a bit

x^4 + 6x^3 - 6x^2 - 45x + 44 = 0

We see x-1 is a factor (do you know how to spot the factors?), so we do long division or synthetic division (whichever one you're better at) and divide x^4 + 6x^3 - 6x^2 - 45x + 44 by x-1 to see that

x^4 + 6x^3 - 6x^2 - 45x + 44 = (x-1)(x^3 + 7x^2 + x - 44) = 0

Now for x^3 + 7x^2 + x - 44 we see that x+4 is a factor. Doing long division again we see that x^3 + 7x^2 + x - 44 = (x+4)(x^2 + 3x - 11)

So finally we have:

(x-1)(x+4)(x^2 + 3x - 11)=0

so x-1 = 0 =>__x=1__

or x+4 = 0 =>__x= -4__

or x^2 + 3x -11 = 0

the latter does not foil nicely, so we can compute by the quadratic formula or completing the square to see that in this case

.......Note the plus or minus sign, this is two answers in one. There are four values for x, as should be expected since we have an x^4 function__x = [-3 +/- sqrt(53)]/2__ - Feb 22nd 2007, 12:33 PMtopsquark
The solution method is correct, but it's not quite that simple. Whenever you square both sides of an equation you might be introducing extra solutions, so you should make a habit of checking all solutions in the original equation.

As it happens, the solution for x is only 1 and -4.

-Dan - Feb 22nd 2007, 01:01 PMSoroban
Hello, Trentt!

This problem is pretty*awful*. .

Quote:

. . . . .__________

x² + √x² + 3x + 5 .= .7 - 3x

We have: .√x² + 3x + 5 .= .7 - 3x - x²

. . . . . . . . . . . . . . . . __________

Square both sides: . (√x² + 3x + 5)² .= .(7 - 3x - x²)²

. . . . x² + 3x + 5 .= .49 - 42x - 5x² + 6x³ + x^4

We have: .x^4 + 6x³ - 6x² - 45x + 44 .= .0

Using the Factor Theorem and Remainder Theorem, I found x = 1 is a root.

. . So we have: .(x - 1)(x³ + 7x² + x - 44) .= .0

I found that x = -4 is a root of the cubic.

. . Then we have: .(x - 1)(x + 4)(x² + 3x - 11) .= .0

- - - . . . . . . . . . . . . . . . . . . . . . . . . . . .__

And the quadratic has roots: .x .= .(-3 ± √53)/2

The answer are: .x .= .1, -4

Neither of the irrational roots satisfies the equation.

. . They are both extraneous.