<s>[INST] Proving $x=\sqrt{a}+\sqrt{b}$ is the key root to solve $x^4-2(a+b)x^2+(a-b)^2=0$ Proving the roots of 

$$x^4-2(a+b)x^2+(a-b)^2=0$$
  are...... 
  $$x=\sqrt{a}+\sqrt{b}$$
  $$x=\sqrt{a}-\sqrt{b}$$
  $$x=-\sqrt{a}+\sqrt{b}$$
  $$x=-\sqrt{a}-\sqrt{b}$$
  When $a$ and $b$ are  real numbers(negative or positive) 

I  proved this by substituting $x=\sqrt{a}+\sqrt{b}$ in the main equation , but I think this way not effective because  It needs to be repeated four times to complete the proving.
Is there another effective proving.  
 [/INST] since
$$(x^2-a-b)^2=4ab\Longrightarrow x^2=a+b\pm 2\sqrt{ab}=(\pm\sqrt{a}\pm \sqrt{b})^2$$
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