# Question 2:

## Part (i):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $16a^2=2\sqrt{a} \Rightarrow a^{\frac{3}{2}}=\frac{1}{8}$ (or equivalent such as $2a^{\frac{1}{2}}(8a^{\frac{3}{2}}-1)=0$) | M1 | Combines two algebraic terms to reach $a^{\pm\frac{3}{2}}=C$ or $(\sqrt{a})^3=C$, $C\neq 0$. Alt: squaring to reach $a^{\pm3}=k$, $k>0$ |
| $a=\left(\frac{1}{8}\right)^{\frac{2}{3}}$ | M1 | Undoes indices correctly for their $a^{\frac{m}{n}}=C$ |
| $a=\frac{1}{4}$ | A1 | $a=\frac{1}{4}$ and no other solutions apart from 0. Accept 0.25 |
| Deduces $a=0$ is a solution | B1 | Must deduce $a=0$ is a solution |

## Part (ii):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $b^4+7b^2-18=0 \Rightarrow (b^2+9)(b^2-2)=0$ | M1 | Attempts to solve as quadratic in $b^2$. Accept $(b^2+m)(b^2+n)=0$ with $mn=\pm18$, or use of quadratic formula with substitution $u=b^2$ |
| $b^2=-9, 2$ | A1 | Correct solution. Allow $b^2=2$ or $u=2$ with no incorrect solution given |
| $b^2=k \Rightarrow b=\sqrt{k},\ k>0$ | dM1 | Finds at least one solution from $b^2=k \Rightarrow b=\sqrt{k}$, $k>0$. Allow $b=1.414$ |
| $b=\sqrt{2},\ -\sqrt{2}$ only | A1 | Only real values; if $3i$ given score A0 |

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