Simplify (1 − 2√(b/a) + b/a) divided by (√a − √b) squared.

Step-by-step solution

1. 1

   Read the numerator as a perfect square. Put $u=\sqrt{b/a}$. Then $b/a=u^{2}$ and the numerator is

   $1-2u+u^{2}=(1-u)^{2}=\left(1-\sqrt{\tfrac{b}{a}}\right)^{2}.$

   Spotting the square pattern $1-2u+u^{2}$ is the whole trick; without it the radicals never cancel.

2. 2

   Rewrite that square over a common denominator. Since $\sqrt{b/a}=\sqrt b/\sqrt a$ for $a>0$,

   $1-\frac{\sqrt b}{\sqrt a}=\frac{\sqrt a-\sqrt b}{\sqrt a},\qquad\text{so}\qquad\left(1-\sqrt{\tfrac ba}\right)^{2}=\frac{(\sqrt a-\sqrt b)^{2}}{a}.$

   Note $(\sqrt a)^{2}=a$, which is where the lone $a$ in the final answer comes from.

3. 3

   Divide by the denominator. The denominator is $\left(a^{1/2}-b^{1/2}\right)^{2}=(\sqrt a-\sqrt b)^{2}$, exactly the factor now sitting in the numerator:

   $\frac{(\sqrt a-\sqrt b)^{2}/a}{(\sqrt a-\sqrt b)^{2}}=\frac{1}{a}.$

4. 4

   State the conditions that make the cancellation legal. We needed $a>0$ for $\sqrt a$ and for $\sqrt{b/a}=\sqrt b/\sqrt a$; $b\ge0$ for $\sqrt b$; and $a\ne b$ so that $(\sqrt a-\sqrt b)^{2}\ne0$ and the division is defined. Under those conditions the value is $1/a$, independent of $b$ entirely.

5. 5

   Test with numbers. Take $a=9$, $b=4$. The numerator is $1-2\sqrt{4/9}+4/9=1-\tfrac43+\tfrac49=\tfrac19$; the denominator is $(3-2)^{2}=1$; the quotient is $\tfrac19=1/a$. With $a=4$, $b=25$: numerator $1-2(2.5)+6.25=2.25$, denominator $(2-5)^{2}=9$, quotient $0.25=1/4$. Both match.