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# Nested Surds

d) $a\sqrt{b}=\sqrt{ab}$

h) $\dfrac{\sqrt{a}+b}{\sqrt{c}+d}=(\sqrt{a}+b)(\sqrt{c}-d)$

i) $\sqrt{5+2\sqrt{6}}=\sqrt{a}+\sqrt{b}$

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### Factorisable Quadratics

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Challenge Level

- Problem
- Student Solutions

For each of the statements provided below, determine which non-negative values of $a$, $b$, $c$, and $d$, if any, make the equation true.

These can be attempted in any order but you might find that some statements can help inform your decisions about others.

You can download these statements as a set of cards that can be cut out and considered in any order.

a) $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$

b) $\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}$

c) $\sqrt{23-6\sqrt{6-4\sqrt{2}}}=\sqrt{a}+\sqrt{b}$

d) $a\sqrt{b}=\sqrt{ab}$

e) $\dfrac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=1$

f) $\sqrt{a} - \sqrt{b}=\sqrt{a - b}$

g) $\sqrt{a}+\sqrt{b}=\sqrt{a+b+\sqrt{4ab}}$

h) $\dfrac{\sqrt{a}+b}{\sqrt{c}+d}=(\sqrt{a}+b)(\sqrt{c}-d)$

i) $\sqrt{5+2\sqrt{6}}=\sqrt{a}+\sqrt{b}$

This comes in two parts, with the first being less fiendish than the second. Itâ€™s great for practising both quadratics and laws of indices, and you can get a lot from making sure that you find all the solutions. For a real challenge (requiring a bit more knowledge), you could consider finding the complex solutions.

You're invited to decide whether statements about the number of solutions of a quadratic equation are always, sometimes or never true.

This will encourage you to think about whether all quadratics can be factorised and to develop a better understanding of the effect that changing the coefficients has on the factorised form.