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