Solve the irrational equation

[Math Tutor]

Solve the irrational equation
[tex]2x^2+3x+3=7x\sqrt{x+1}[/tex]

It is difficult!

[Rock'n'roller]

Let x = a² - 1

We have

[tex]2(a^2-1)^2 + 3(a^2-1) + 3 = 7a(a^2-1) or[/tex]

[tex]2a^4 - 7a^3 - a^2 + 7a + 2 = 0[/tex]

[tex](2a^2 + ma + n)(a^2 + pa + q) = 0[/tex]

[tex]2a^4 + (m+2p)a^3 + (n + mp + 2q)a^2 + (np + mq)a + nq = 0[/tex]

Then we solve the system

m + 2p = -7
n + mp + 2q = -1
np + mq = 7
nq = 2

And then it's easy

[HallsofIvy]

I confess that wouldn't have occurred to me! I would have done the obvious- square both sides.

$(2x^2+ 3x+ 3)^2= 2x^2(2x^2+ 3x+ 3)+ 3x(2x^2+ 3x+ 3)+ 3(2x^2+ 3x+ 3)= 4x^4+ 6x^3+ 6x^2+ 6x^3+ 9x^2+ 9x+ 6x^2+ 9x+ 9= 4x^4+ 12x^3+ 21x^2+ 18x+ 9= 49x^2(x- 1)= 49x^3- 49x$.

$4x^4- 37x^3+ 21x^2- 31x+ 9= 0$.

That is a fourth degree equation which can be very difficult to solve. I would now try the "rational root theorem" which says that **if** there is rational root (there is no guarantee that the there is one) then it is of the form [tex]\frac{p}{q}[/tex] where p divides the "leading coefficient", 4, and p divides the "constant term", 9.

So p must be one of 1, -1, 3, -3, 9, or -9 and q must be one of 1, -1, 2, -2, 4, or -4. The only possible rational roots are 1, -1, 3, -3, 9, -9, 1/2, -1/2, 1/4, -1/4, 3/2, -3/2, 3/4, -3/4, 9/2, 9/2, 9/4, or -9/4. Evaluate the polynomial at each of those.