Indices, surds and factorising

Mark scheme

1. Answer. $2$

   • $16^{3/4} = (2^4)^{3/4} = 2^3 = 8$.
   • $8^{2/3} = (2^3)^{2/3} = 2^2 = 4$.
   • $\dfrac{8}{4} = 2$.

2. Answer. $5\sqrt{3}$

   • $\sqrt{48} = 4\sqrt{3}$, $\sqrt{12} = 2\sqrt{3}$ and $\sqrt{27} = 3\sqrt{3}$.
   • $4\sqrt{3} - 2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$.

3. Answer. $x^2 + 7x - 13$

   • $(2x - 3)(x + 4) = 2x^2 + 5x - 12$.
   • $(x - 1)^2 = x^2 - 2x + 1$.
   • Subtract to get $x^2 + 7x - 13$.

4. Answer. $(3x - 4)(2x + 3)$

   • Look for two numbers whose product is $6 \times (-12) = -72$ and whose sum is $1$: $9$ and $-8$.
   • Split the middle term and factorise to get $(3x - 4)(2x + 3)$.

5. Answer. $\dfrac{3}{2} - \dfrac{1}{2}\sqrt{5}$

   • Multiply the numerator and denominator by $\sqrt{5} - 1$.
   • The numerator is $(\sqrt{5} - 1)^2 = 6 - 2\sqrt{5}$.
   • The denominator is $4$.
   • $\dfrac{6 - 2\sqrt{5}}{4} = \dfrac{3}{2} - \dfrac{1}{2}\sqrt{5}$.

6. Answer. $x = 2$

   • Write both sides as powers of $2$: $2^{3x} = 2^{2x + 2}$.
   • $3x = 2x + 2$.
   • $x = 2$.

7. Answer. $x = 0$ or $x = 2$

   • Let $y = 2^x$. Then $y^2 - 5y + 4 = 0$.
   • $(y - 4)(y - 1) = 0$, so $y = 4$ or $y = 1$.
   • $2^x = 4$ gives $x = 2$, and $2^x = 1$ gives $x = 0$.

8. Answer. $2$

   • $\sqrt{8} = 2\sqrt{2}$, so the expression is $\dfrac{4 + 2\sqrt{2}}{2 + \sqrt{2}}$.
   • Multiply the numerator and denominator by $2 - \sqrt{2}$.
   • The numerator is $(4 + 2\sqrt{2})(2 - \sqrt{2}) = 4$.
   • The denominator is $2$, so the value is $2$.