Algebra · real student question

Simplify (√a + 1)/(a − 1).

Question

Simplify

a+1a1(a0, a1)\frac{\sqrt a+1}{a-1}\qquad(a\ge 0,\ a\neq 1)

Step-by-step solution

  1. Spot the hidden difference of squares. For a0a\ge 0 we have a=(a)2a=\left(\sqrt a\right)^{2} and 1=121=1^{2}, so the denominator factors:

    a1=(a)212=(a1)(a+1)a-1=\left(\sqrt a\right)^{2}-1^{2}=\left(\sqrt a-1\right)\left(\sqrt a+1\right)

    This is the key move — without it, nothing in the fraction looks cancellable.

  2. Rewrite the fraction with the factored denominator.

    a+1(a1)(a+1)\frac{\sqrt a+1}{\left(\sqrt a-1\right)\left(\sqrt a+1\right)}

  3. Cancel the common factor. Since a0a\ge 0, a+11>0\sqrt a+1\ge 1>0, so the cancellation is always legitimate — no extra exclusion is introduced:

    1a1\frac{1}{\sqrt a-1}

  4. Record the domain. The original expression needs a0a\ge 0 (for the square root) and a1a\neq 1 (denominator nonzero); the simplified form carries the same two restrictions, since a1=0\sqrt a-1=0 exactly when a=1a=1.

    1a1(a0, a1)\boxed{\dfrac{1}{\sqrt a-1}\qquad(a\ge 0,\ a\neq 1)}

  5. Check with a value. At a=9a=9: the original is 3+191=48=0.5\dfrac{3+1}{9-1}=\dfrac48=0.5, and the simplified form gives 131=0.5\dfrac{1}{3-1}=0.5 ✓. At a=0.25a=0.25: original =0.5+10.251=1.50.75=2=\dfrac{0.5+1}{0.25-1}=\dfrac{1.5}{-0.75}=-2, simplified =10.51=2=\dfrac{1}{0.5-1}=-2 ✓ — including the negative branch 0a<10\le a<1.

Answer

1a1(a0, a1)\dfrac{1}{\sqrt a-1}\qquad(a\ge 0,\ a\neq 1)

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