A standard surd-manipulation problem, not tied to a specific exam paper.
Problem
Find the positive square root of $3+\sqrt5$.
Key insight. A “nested” square root like $\sqrt{3+\sqrt5}$ can sometimes be un-nested into a sum of two simpler square roots, $\sqrt{x}+\sqrt{y}$. Squaring that assumed form produces both a rational part ($x+y$) and an irrational part ($2\sqrt{xy}$) — matching each separately against the original expression’s rational and irrational parts gives two equations that pin down $x$ and $y$.
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Approach
Assume $\sqrt{3+\sqrt5} = \sqrt{x}+\sqrt{y}$ for some $x,y$. Square both sides, then match the rational and irrational parts separately to get two equations in $x$ and $y$. Solve this system using the identity linking $(x-y)^2$ to $(x+y)^2$ and $xy$.
Solution
Step 1 — Assume the un-nested form
$$\sqrt{3+\sqrt5} = \sqrt{x}+\sqrt{y}$$
Step 2 — Square both sides
$$3+\sqrt5 = x+y+2\sqrt{xy}$$
Step 3 — Match rational and irrational parts
Comparing the rational parts: $x+y=3$
Comparing the irrational (√5) parts: $2\sqrt{xy}=\sqrt5$
Step 4 — Solve for xy
Squaring the second equation:
$$4xy=5 \implies xy=\frac{5}{4}$$
Step 5 — Find x−y using the identity
$$(x-y)^2 = (x+y)^2-4xy = 9-5=4 \implies x-y=\pm2$$
Taking the positive value (since the positive square root is required):
$$x-y=2$$
Step 6 — Solve the linear system
Adding $x+y=3$ and $x-y=2$:
$$2x=5 \implies x=\frac{5}{2}, \qquad y = 3-\frac{5}{2} = \frac{1}{2}$$
Step 7 — Write the final answer
$$\sqrt{3+\sqrt5} = \sqrt{\frac{5}{2}}+\sqrt{\frac{1}{2}}$$
Answer
$$\sqrt{\frac{5}{2}}+\sqrt{\frac{1}{2}} = \frac{\sqrt{10}+\sqrt2}{2}$$
Common mistakes
- Forgetting to square the equation $2\sqrt{xy}=\sqrt5$ correctly. Squaring gives $4xy=5$, not $2xy=5$ — dropping the factor of $4$ from the squared $2\sqrt{xy}$ leads to the wrong value of $xy$.
- Choosing $x-y=-2$ instead of $+2$. Both satisfy the squared equation, but only $x-y=+2$ (assuming $x>y$ as set up here) leads to the positive square root asked for; the other sign would just swap $x$ and $y$, giving the same final answer here, but in general the sign choice matters for which root (positive or negative) is being found.
Practise next
- Find the positive square root of $7+\sqrt{24}$, using the same un-nesting method.
Show answer
$1+\sqrt6$. Write $\sqrt{24}=2\sqrt6$, so the expression is $7+2\sqrt6$.
Look for $\left(\sqrt m+\sqrt n\right)^2=m+n+2\sqrt{mn}$ with $m+n=7$ and $mn=6$ — that is $m,n=1,6$.
So $7+2\sqrt6=\left(1+\sqrt6\right)^2$ and the positive square root is $1+\sqrt6\approx3.449$. The trick only works when $m+n$ and $mn$ come out as integers, which is why the surd has to be put in the form $2\sqrt{mn}$ first.
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