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.
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Calculate the determinant:$$4(a+b)^2-4(a-b)^2 = 16ab$$ Then find the roots as following: $$x^2 = \frac{2(a+b)+\sqrt{16ab}}{2},\frac{2(a+b)-\sqrt{16ab}}{2}$$ $$x^2 = a+b+2\sqrt {ab}, a+b-2\sqrt {ab}$$ $$x = \pm(\sqrt a + \sqrt b), \pm(\sqrt a - \sqrt b)$$
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One way to verify this with minimal computations is to expand the product $P(x) = \prod (x\pm\sqrt{a}\pm\sqrt{b})$ (where the product runs over all combinations of signs).
Taking as a variable $t_1 = x+\sqrt{a}$ and $t_2 = x-\sqrt{a}$, this becomes
$$P(x) = (t_1 + \sqrt{b})(t_1-\sqrt{b})(t_2 + \sqrt{b})(t_2-\sqrt{b}) = (t_1^2 - b)(t_2^2 - b).$$
Now $t_1t_2 = x^2-a$ and $t_1^2 + t_2^2 = 2(x^2 + a)$. So
$$P(x) = (x^2-a)^2 - 2b(x^2+a) + b^2 = x^4 - 2(a+b)x^2 +a^2 - 2ab + b^2. $$
This being said, the from math110's answer is nicer. In fact if you read it like this, it is even more obvious.
$$ \begin{aligned} x = \sqrt{a} + \sqrt{b} & \overset{\mathrm{square}}\implies x^2 = a^2 + b^2 + 2\sqrt{ab} \\ &\overset{\mathrm{rewrite}}\implies x^2 - a^2 - b^2 = 2\sqrt{ab} \\ &\overset{\mathrm{square}}\implies (x^2-a^2-b^2)^2 = 4ab. \end{aligned}$$
If $r$ and $s$ are solutions to your equation, then we consider $r^2$ and $s^2$: $$r^2+s^2=2(a+b)$$ and $$r^2s^2=(a-b)^2$$ These simultaneous equations can be solved fairly easily, giving us two similar formulas, with a $\pm$, for $r^2$ and $s^2$. Taking square roots gives your four solutions, with another $\pm$. All this shows why the solutions are so similar in form, with two plus-or-minuses.