Thread: Radicals Problem

x² + root(x²+3x+5) = 7 - 3x


Thanks

Originally Posted by Trentt

x² + root(x²+3x+5) = 7 - 3x


Thanks

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
x = [-3 +/- sqrt(53)]/2 .......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

Originally Posted by Jhevon

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
x = [-3 +/- sqrt(53)]/2 .......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

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

Hello, Trentt!

This problem is pretty awful . .

. . . . .__________
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.