If √(a + √b) = 3 + √2, Then Find √(a² − b)

A standard quadratic equations problem, not tied to a specific exam paper.

Problem

If $\sqrt{a+\sqrt b}=3+\sqrt2$, then find $\sqrt{a^2-b}$.

Note: the worked solution embedded below states the equation as $\sqrt{a-\sqrt b}=3+\sqrt2$. The values it derives, $a=11$ and $b=72$, satisfy only the plus-sign form, so that form is used here. The answer, $7$, is the same either way.

Key insight. Squaring the given equation produces an expression of the form (rational) $+$ (multiple of $\sqrt2$) — matching that against $a+\sqrt b$ pins down $a$ as the rational part, and matching the $\sqrt2$ coefficient against $\sqrt b$ pins down $b$, all without needing to solve two separate equations.

Watch this explained step by step →  ·  More on the Shiwam’s Classes channel

Approach

Square both sides of $\sqrt{a+\sqrt b}=3+\sqrt2$ to remove the outer square root, expanding the right side fully. Match the resulting rational part to $a$ and the coefficient of $\sqrt2$ to $\sqrt b$ (squaring that coefficient, since $\sqrt b$ itself is what’s being matched). With $a$ and $b$ both known, compute $a^2-b$ directly and take its square root.

Solution

Step 1 — Square both sides

$$a+\sqrt b = (3+\sqrt2)^2$$

Step 2 — Expand the right side

$$(3+\sqrt2)^2 = 9+6\sqrt2+2 = 11+6\sqrt2$$

Step 3 — Match rational and irrational parts

$$a+\sqrt b = 11+6\sqrt2$$

Matching the rational part: $a=11$. Matching the coefficient of $\sqrt2$: $\sqrt b$ corresponds to $6\sqrt2$, so $b=(6\sqrt2)^2=36\times2=72$.

Step 4 — Compute a² − b

$$a^2-b = 11^2-72 = 121-72 = 49$$

Step 5 — Take the square root

$$\sqrt{a^2-b} = \sqrt{49} = 7$$

Answer

Common mistakes

• Forgetting to square the coefficient of $\sqrt2$ when finding $b$. Since $\sqrt b$ corresponds to $6\sqrt2$, $b=(6\sqrt2)^2=72$ — using $b=6\times2=12$ (forgetting to square the whole term) badly understates $b$.
• Expanding $(3+\sqrt2)^2$ incorrectly. The middle term of the expansion is $2\times3\times\sqrt2=6\sqrt2$, not $3\sqrt2$ or $9\sqrt2$ — dropping the factor of $2\times3$ is a common slip.
• Computing $a^2-b$ using $a=11$ and the unsquared coefficient 6 instead of $b=72$. The question asks for $a^2-b$, which needs the actual value of $b$ (a plain number), not the surd coefficient that led to it.

Practise next

• If $\sqrt{a+\sqrt b}=2+\sqrt5$, find $\sqrt{a^2-b}$.