Solve the equation without using square both sides

Question

Solve the equation without using square both sides:

$$(x+1)(2x^{3}-5)^{1/2}= 2x^{3}- 3x^{2}+x-2$$

I try to use Wolfram Alpha and here is the the answer:

I can' t continue. Help me!

because the degree will be too high and the solution will have some complex or superfluous roots, I don' t like this! — Đăng Đặng Hai, Apr 15 '18 at 08:45
I think the equation just like $(A+\sqrt{B})( C+\sqrt{D} )= 0$ — Đăng Đặng Hai, Apr 15 '18 at 08:49

sirous · Answer 1 · 2018-04-15T18:11:04.250

-1

The solution is $x=3$ and can be found as follows:

$2 x^3-5$ must be perfect square among cubic of few numbers we only have:

$2\times 3^3-5=54-5=49=7^2$

We check this:

$(3+1)\times 7= 28$

$2\times3^3-3\times 3^2+3-2=28$

Now suppose $(2x^3-5)^{1/2}=a$ then we have:

$a(x+1)=a^2-3x^2+x+3$.. or $a^2-(x+1)a-3x^2+x+3=0$

The condition for having solution in R is:

$\Delta=13x^2-2x-11=m^2; m ∈ R$

$2x^3-5=k^2; k∈ R$

The solution of question leads to the solution of this Diophantine system of equations.One solution is $x=3,..k=7,..m=10$

edited Apr 15 '18 at 18:11

answered Apr 15 '18 at 09:43

sirous

I' m so sorry. There can' t only be integer roots. The equation also have the real roots. – Đăng Đặng Hai Apr 15 '18 at 09:49

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