Splitting a Single Eqn Into Two

Answer to the Bonus question :

Given $\sqrt{x + a} + \sqrt{x+b} = c$ . Let $\sqrt{x+a} - \sqrt{x+b} = d$ .

Using the difference of squares identity, $A^2 - B^2 = C^2$ , multiplying the two equations above gives $a - b = cd \Rightarrow d = \frac{a-b}c$ .

Now adding the two equations $\begin{cases} \sqrt{x + a} + \sqrt{x+b} = c \\ \sqrt{x + a} - \sqrt{x+b} = \frac{a-b}c \end{cases}$ gives $2 \sqrt{x+a} = c + \frac{a-b}c \Rightarrow x = \left(\frac{c + \frac{a-b}c}2 \right)^2 - a .$

the trick is, as the title suggests, to write 2 eqns. $\sqrt{x+25}+\sqrt{x+49}=18 \to (\sqrt{x+25}+\sqrt{x+49})(\sqrt{x+25}-\sqrt{x+49})= 18 (\sqrt{x+25}-\sqrt{x+49}) \\ \to\sqrt{x+25}-\sqrt{x+49}=\dfrac{x+25-(x+49)}{18} = -\dfrac{4}{3}$ we have the two eqns $\sqrt{x+25}+\sqrt{x+49}=18 \\\sqrt{x+25}-\sqrt{x+49}=-\dfrac{4}{3}$ adding the two eqns we get $2\sqrt{x+25} = 18-\dfrac{4}{3} \to x = \boxed{\dfrac{400}{9}}$ pluging this back into the eqn, we confirm it works.

The problem is obviously symmetric in $a, b$ , so the answer should be a function of their sum and difference, $s = \tfrac12(a+b)$ and $d = \tfrac12|a-b|$ .

Write $a = s + d$ and $b = s - d$ . The equation becomes $c = \sqrt{x + s + d} + \sqrt{x + s - d}.$ Similar to Pi Han Goh's approach, we define $q := \sqrt{x + s + d} - \sqrt{x + s - d},$ and observe that $cq = (x+s+d) - (x+s-d) = 2d\ \ \ \therefore\ \ \ q = \frac{2d}{c};$ Adding and subtracting the equations shows that $c \pm \frac{2d}{c} = 2\sqrt{x + s \pm d}.$ From this we find that $x = \left(\frac{c}{2} \pm \frac{d}{c}\right)^2 - s \mp d,$ $= \left(\frac{c}{2}\right)^2 + \left(\frac{d}{c}\right)^2 \pm d - s \mp d$ $= \left(\frac{c}{2}\right)^2 + \left(\frac{d}{c}\right)^2 - s.$ In this case, $d = \tfrac12(49-25) = 12$ and $s = \tfrac12(49 + 25) = 37$ , so that $x = 9^2 + (\tfrac 23)^2 - 37 = 44\tfrac 4 9 = \frac{400}{9}.$

$\sqrt{x + a} + \sqrt{x + b} = c$

$(\sqrt{x + a} + \sqrt{x + b})^2 = c^2$

$(x + a) + 2\sqrt{(x + a)(x + b)} + (x + b) = c^2$

$2\sqrt{(x + a)(x + b)} = c^2 - (x + a) - (x + b)$

$(2\sqrt{(x + a)(x + b)})^2 = (c^2 - (x + a) - (x + b))^2$

$4(x + a)(x + b) = (c^2 - (x + a) - (x + b))^2$

which simplifies to $x = \frac{(c^2 + a - b)^2}{4c^2} - a$ or equivalently $x = \frac{(c^2 - a + b)^2}{4c^2} - b$ .

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In the question, $a = 25$ , $b = 49$ , $c = 18$ , so $x = \frac{(18^2 + 25 - 49)^2}{4 \cdot 18^2} - 25 = \frac{400}{9}$ , so $A = 400$ , $B = 9$ , and $\sqrt{A} + \sqrt{B} = \boxed{23}$ .

There are already great solutions to this so I'm not adding much, but wanted to highlight that there is a slightly neater way to clear the square roots.

If we square both sides of the equation as it is, we get a cross-term involving the product of two $x$ -terms, and after rearranging and squaring again, end up with a quadratic in $x$ . Everything cancels nicely here, but there's no need for the quadratic to appear at all.

Rearranging at the start to

$\sqrt{x+a}=c-\sqrt{x+b}$

and then squaring, we have

$x+a=c^2-2c\sqrt{x+b}+x+b$

Which simplifies to

$2c\sqrt{x+b}=c^2-a+b$

Having already cancelled some $x$ s, this is now a bit simpler to solve:

$x=\frac{(c^2-a+b)^2}{4c^2}-b=\boxed{\frac{a^2+b^2+c^4-2ab-2ac^2-2bc^2}{4c^2}}$

(the last form shows the symmetry in $a$ and $b$ a bit more clearly.)

Solving $\sqrt{a+x}+\sqrt{b+x}=c$ for $x$ gives $\frac{a^2-2 a b-2 a c^2+b^2-2 b c^2+c^4}{4 c^2}$ . Substituting $a\to 25$ , $b\to 49$ and $c\to 18$ gives $x$ as $\frac{400}{9}$ . $\sqrt{\text{Denominator}[\frac{400}{9}]}+\sqrt{\text{Numerator}[\frac{400}{9}]} \Rightarrow 23$ .