Topics › KS4 › Number · Higher only

Surds

1. surd-01 Present from here

   Write $\sqrt{48}$ in its simplest form.

   A Option A: $16\sqrt{3}$

   B Option B: $8\sqrt{3}$

   C Option C: $4\sqrt{3}$

   D Option D: $3\sqrt{16}$
   Answer and misconceptions

   A Option A: $16\sqrt{3}$ ✗

   Moves the square factor outside without square-rooting itWrites $\sqrt{16 \times 3}$ as $16\sqrt{3}$; the $16$ must be square-rooted on its way out: $\sqrt{16} = 4$.

   B Option B: $8\sqrt{3}$ ✗

   Halves the square factor instead of square-rooting itTakes $\sqrt{16}$ as $8$; the square root of $16$ is $4$, because $4 \times 4 = 16$.

   C Option C: $4\sqrt{3}$ ✓ Correct

   D Option D: $3\sqrt{16}$ ✗

   Puts the non-square factor outside the rootWrites $\sqrt{16 \times 3}$ as $3\sqrt{16}$, which equals $12$. The square factor $16$ is the one that comes out, as $4$.
2. surd-02 Present from here

   Work out $2\sqrt{3} \times 4\sqrt{3}$

   A Option A: $24$

   B Option B: $8\sqrt{3}$

   C Option C: $6\sqrt{3}$

   D Option D: $72$
   Answer and misconceptions

   A Option A: $24$ ✓ Correct

   B Option B: $8\sqrt{3}$ ✗

   Thinks $\sqrt{3} \times \sqrt{3} = \sqrt{3}$Multiplies $2 \times 4 = 8$ but keeps one $\sqrt{3}$; $\sqrt{3} \times \sqrt{3} = 3$, so the answer is $8 \times 3$.

   C Option C: $6\sqrt{3}$ ✗

   Adds the surds instead of multiplyingWorks out $2\sqrt{3} + 4\sqrt{3} = 6\sqrt{3}$; the question asks for a product.

   D Option D: $72$ ✗

   Thinks $\sqrt{3} \times \sqrt{3} = 9$Squares the $3$ instead of seeing that $\sqrt{3} \times \sqrt{3} = (\sqrt{3})^2 = 3$.
3. surd-03 Present from here

   Rationalise the denominator of $\frac{6}{\sqrt{2}}$ and simplify.

   A Option A: $6\sqrt{2}$

   B Option B: $3$

   C Option C: $\frac{3\sqrt{2}}{2}$

   D Option D: $3\sqrt{2}$
   Answer and misconceptions

   A Option A: $6\sqrt{2}$ ✗

   Multiplies only the numerator by $\sqrt{2}$Multiplies the top by $\sqrt{2}$ but not the bottom, which changes the value. Multiply by $\frac{\sqrt{2}}{\sqrt{2}}$ to get $\frac{6\sqrt{2}}{2}$.

   B Option B: $3$ ✗

   Treats $\sqrt{2}$ as $2$Works out $6 \div 2 = 3$; $\sqrt{2}$ is about $1.41$, not $2$.

   C Option C: $\frac{3\sqrt{2}}{2}$ ✗

   Thinks $\sqrt{2} \times \sqrt{2} = 4$Gets a denominator of $4$; $\sqrt{2} \times \sqrt{2} = 2$, so the fraction is $\frac{6\sqrt{2}}{2} = 3\sqrt{2}$.

   D Option D: $3\sqrt{2}$ ✓ Correct
4. surd-04 Present from here

   Simplify $\sqrt{50} + \sqrt{18}$

   A Option A: $\sqrt{68}$

   B Option B: $8\sqrt{2}$

   C Option C: $8\sqrt{4}$

   D Option D: $34\sqrt{2}$
   Answer and misconceptions

   A Option A: $\sqrt{68}$ ✗

   Thinks $\sqrt{a} + \sqrt{b} = \sqrt{a + b}$Adds the numbers under the roots. Simplify each surd first: $5\sqrt{2} + 3\sqrt{2}$.

   B Option B: $8\sqrt{2}$ ✓ Correct

   C Option C: $8\sqrt{4}$ ✗

   Adds the numbers under the roots as well as the coefficientsWrites $5\sqrt{2} + 3\sqrt{2} = 8\sqrt{4}$. Like surds add like algebraic terms: $5$ lots of $\sqrt{2}$ plus $3$ lots of $\sqrt{2}$ is $8\sqrt{2}$.

   D Option D: $34\sqrt{2}$ ✗

   Moves square factors outside without square-rooting themWrites $\sqrt{25 \times 2}$ as $25\sqrt{2}$ and $\sqrt{9 \times 2}$ as $9\sqrt{2}$; they should be $5\sqrt{2}$ and $3\sqrt{2}$.
5. surd-05 Present from here

   Expand and simplify $(3 + \sqrt{5})(3 - \sqrt{5})$

   A Option A: $4$

   B Option B: $14$

   C Option C: $-16$

   D Option D: $9 - \sqrt{5}$
   Answer and misconceptions

   A Option A: $4$ ✓ Correct

   B Option B: $14$ ✗

   Thinks $\sqrt{5} \times (-\sqrt{5}) = +5$Gets the sign wrong on the last term. A positive times a negative is negative, so the product is $9 - 5$.

   C Option C: $-16$ ✗

   Thinks $\sqrt{5} \times \sqrt{5} = 25$Squares the $5$; $\sqrt{5} \times \sqrt{5} = 5$, so the last term is $-5$, not $-25$.

   D Option D: $9 - \sqrt{5}$ ✗

   Thinks $\sqrt{5} \times \sqrt{5} = \sqrt{5}$Keeps the root after multiplying; $\sqrt{5} \times \sqrt{5} = 5$.