Aram's real solutions

Let us rewrite $\sqrt{(x+1)(x-1)^2} - 2x\sqrt{x-2} = 0$ the following way:

Step 1: $\sqrt{(x+1)(x-1)^2} = 2x\sqrt{x-2}$

Step 2 : We square both sides to get - $(x+1)(x-1)^3 = 4x^{2}(x-2)$

Step 3: We open up the brackets to get - $x^4-6x^3+8x^2+2x-1=0$

Step 4: The idea here is to factorise the above equation. Well to do this, we can remark that $-6 = -4 + -2$ , $+8 = -4 \times -2$ and (coincidentally) $+2 = (-4) - (-2)$ . Now this motivates us to write: $x^4-6x^3+8x^2+2x-1= (x^2 - 4x + 1) (x^2 - 2x -1)$

Step 5: Now we need to solve:

(a) $x^2 - 4x + 1 = 0$ or

(b) $x^2 - 2x -1 =0$

A reasonable prediction would be 4 roots. However due to the details and assumption, we can in fact directly obtain from the question the important condition that $x ≥ 2$ . This implies that for:

(a) We need $(x-2)^2 = 3$ We get $x = 2 - \sqrt{3}$ or $2 + \sqrt{3}$ We can reject the first root since it is less than 2.

(b) Similar analysis as the above case shows that we can reject the second root since: $1 - \sqrt{2} < 1 < 2$ . So for this case we get one root namely: $1 + \sqrt{2}$ .

In conclusion the answer is : $1 + \sqrt{2} + 2 + \sqrt{3} = \sqrt{9} + \sqrt{2} + \sqrt{3}$ Hence _9+2+3 = 14 _

Any solution of the equation $\sqrt{(x+1)(x-1)^{3}} - 2x\sqrt{x-2} = 0$ must be such that $x ≥ 2$ .

$\sqrt{(x+1)(x-1)^{3}} - 2x\sqrt{x-2} = 0$

$\sqrt{(x+1)(x-1)^{3}} = 2x\sqrt{x-2}$

$(x+1)(x-1)^{3} = 4x^{2}(x-2)$

$(x+1)(x^{3}-3x^{2}+3x-1) = 4x^{2}(x-2)$

$x^{4}-2x^{3}+2x-1=4x^{3}-8x^{2}$

$x^{4}-6x^{3}+8x^{2}+2x-1=0$

$(x^{2}-2x-1)(x^{2}-4x+1)=0$

$x=1+\sqrt{2}$ or $x=2+\sqrt{3}$

Let $x_{1}=1+\sqrt{2}$ and $x_{2}=2+\sqrt{3}$ so that the sum of all real solutions is $x_{1}+x_{2}=3+\sqrt{2}+\sqrt{3}=\sqrt{9}+\sqrt{2}+\sqrt{3}$ . Let $a=9$ , $b=2$ , and $c=3$ so that $a+b+c=9+2+3=14$ .

$\sqrt{(x+1^) *(x-1)}^{3} = 2x \sqrt{x-2}$

$(x+1)(x-1)^{3} = 4x^{3} - 8x^{2}$

$x^{4} - 2x^{3} + 2x - 1$ = $4x^{3} - 8x^{2}$

$x^{4} - 6x^{3} + 8x^{2} + 2x - 1 = 0$

Let $q,p,r,s$ be the roots. From Vieta's Formulas we know that in a forth degree polynomial. $(-p-q-r-s)$ = coefficient of $x^{3}$

$p + q + r + s = 6$

Finally find three diferent numbers $a,b,c$ with $a + b + c = 6.$

They are $1 , 2 , 3$ . So $\sqrt{1} + \sqrt{4} + \sqrt{9}$ = $6$

So $1 + 4 + 9 = 14$

I don't see why this is rated so highly. All you have to do is expand, simplify, and then you'll get a parabola. It has roots at $1+\sqrt{2}$ and $2+\sqrt{3}$ . Adding the roots gives $\sqrt2+\sqrt3+\sqrt9$ . The desired sum is thus $2+3+9=\boxed{14}$ .