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# Nested Surds

David from Dinnington High School has got us started with a solution to f) $\sqrt{a}-\sqrt{b} =\sqrt{a-b}$

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

$\implies (\sqrt{a}-\sqrt{b})^2 =a-b$

$\implies a-2\sqrt{a b}+b=a-b$

$\implies 2b-2\sqrt{a b} =0$

$\implies b =\sqrt{a b}$

$\implies b = \sqrt{a} \sqrt{b}$

$\implies \frac{b}{\sqrt{b}} =\sqrt{a}$

$\implies \sqrt{b} =\sqrt{a}$

$\implies a=b$

What this shows is that if $a$ and $b$ are going to satisfy $\sqrt{a}-\sqrt{b}=\sqrt{a-b}$ then $a$ must be equal to $b$. In fact part of the solution has been missed because in the 7th line we divided by $\sqrt{b}$ which assumed $b \neq 0$, so in the fact the possibilites are that $a=b$ or $b=0$. Conversely we can check that if $a=b$ or $b=0$ then $a$ and $b$ satisfy $\sqrt{a}-\sqrt{b}=\sqrt{a-b}$. So in summary $\sqrt{a}-\sqrt{b} =\sqrt{a-b}$ if and only if $a=b$ or $b=0$.

Sergio from King's College of Alicante has found a solution to i) $\sqrt{5+2 \sqrt{6}} = \sqrt{a}+\sqrt{b}$

In the first place I decided to square everything to get rid of some square roots:

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

Now I looked carefully and I noticed that I probably had found two equations, so I now needed to solve simultaneous equations!

$5=a+b$

$6=ab$

This means that

$6=b(5-b)$

$0=b^2-5b+6$

I solved it by factorising, although you could have also used the quadratic equation, which gave me that $b$, and also $a$ when I substituted in, could be either 3 or 2.

Thomas from BHASVIC Sixth Form College Brighton has solved the remaining parts:

a) Squaring both sides of the equation we simply get $a \times b = ab$ which is clearly true for all non-negative $a$ and $b$.

b) Again, squaring both sides gives $\frac{a}{b} = \frac{a}{b}$, which again is true for all non-negative $a$ and $b>0$.

c) The largest value of $b$ occurs at $a=0$, when $b=23-6\sqrt{6-4\sqrt{2}}$. As $a$ increases, $b$ decreases until it reaches zero, and $a=23-6\sqrt{6-4\sqrt{2}}$. Squaring gives the solution set $a=(\sqrt{23-6\sqrt{6-4\sqrt{2}}}-\sqrt{b})^2$ for $0 \leq b \leq 23-6\sqrt{6-4\sqrt{2}}$.

d) Squaring both sides gives $a^2b=ab \implies ab(a-1) = 0$. So have solutions $a=0,1$ for all non-negative $b$, and $b=0$ for all non-negative $a$.

e) Divide the top and bottom of the fraction $\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}$ by $\sqrt{a}$. Then multiply through by denominator and rearrange to isolate $\sqrt{a}$:

$$\sqrt{a}=\frac{\sqrt{b}}{\sqrt{b}-1}$$

The LHS is positive, so must have denominator $\sqrt{b}-1>0$, so $b>1$. So solutions are given by

$$a = \frac{b}{(\sqrt{b}-1)^2}$$

for $b>1$.

g) Squaring both sides of the equation gives $$a+b+2\sqrt{ab}=a+b+\sqrt{4ab}$$ which is clearly true for all non-negative $a$ and $b$.

h) Take the RHS to the LHS of the equation and factorise out $\sqrt{a}+b$, so that

$$(\sqrt{a}+b)(\frac{c-d^2}{\sqrt{c}+d})=0$$

Then have the solution sets:

$a=b=0$ for any $c$ and $d$ non-negative not both equal to zero

or

any non-negative $a,b$ and all positive $c=d^2$.

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Age 16 to 18

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- Student Solutions

David from Dinnington High School has got us started with a solution to f) $\sqrt{a}-\sqrt{b} =\sqrt{a-b}$

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

$\implies (\sqrt{a}-\sqrt{b})^2 =a-b$

$\implies a-2\sqrt{a b}+b=a-b$

$\implies 2b-2\sqrt{a b} =0$

$\implies b =\sqrt{a b}$

$\implies b = \sqrt{a} \sqrt{b}$

$\implies \frac{b}{\sqrt{b}} =\sqrt{a}$

$\implies \sqrt{b} =\sqrt{a}$

$\implies a=b$

What this shows is that if $a$ and $b$ are going to satisfy $\sqrt{a}-\sqrt{b}=\sqrt{a-b}$ then $a$ must be equal to $b$. In fact part of the solution has been missed because in the 7th line we divided by $\sqrt{b}$ which assumed $b \neq 0$, so in the fact the possibilites are that $a=b$ or $b=0$. Conversely we can check that if $a=b$ or $b=0$ then $a$ and $b$ satisfy $\sqrt{a}-\sqrt{b}=\sqrt{a-b}$. So in summary $\sqrt{a}-\sqrt{b} =\sqrt{a-b}$ if and only if $a=b$ or $b=0$.

Sergio from King's College of Alicante has found a solution to i) $\sqrt{5+2 \sqrt{6}} = \sqrt{a}+\sqrt{b}$

In the first place I decided to square everything to get rid of some square roots:

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

Now I looked carefully and I noticed that I probably had found two equations, so I now needed to solve simultaneous equations!

$5=a+b$

$6=ab$

This means that

$6=b(5-b)$

$0=b^2-5b+6$

I solved it by factorising, although you could have also used the quadratic equation, which gave me that $b$, and also $a$ when I substituted in, could be either 3 or 2.

Thomas from BHASVIC Sixth Form College Brighton has solved the remaining parts:

a) Squaring both sides of the equation we simply get $a \times b = ab$ which is clearly true for all non-negative $a$ and $b$.

b) Again, squaring both sides gives $\frac{a}{b} = \frac{a}{b}$, which again is true for all non-negative $a$ and $b>0$.

c) The largest value of $b$ occurs at $a=0$, when $b=23-6\sqrt{6-4\sqrt{2}}$. As $a$ increases, $b$ decreases until it reaches zero, and $a=23-6\sqrt{6-4\sqrt{2}}$. Squaring gives the solution set $a=(\sqrt{23-6\sqrt{6-4\sqrt{2}}}-\sqrt{b})^2$ for $0 \leq b \leq 23-6\sqrt{6-4\sqrt{2}}$.

d) Squaring both sides gives $a^2b=ab \implies ab(a-1) = 0$. So have solutions $a=0,1$ for all non-negative $b$, and $b=0$ for all non-negative $a$.

e) Divide the top and bottom of the fraction $\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}$ by $\sqrt{a}$. Then multiply through by denominator and rearrange to isolate $\sqrt{a}$:

$$\sqrt{a}=\frac{\sqrt{b}}{\sqrt{b}-1}$$

The LHS is positive, so must have denominator $\sqrt{b}-1>0$, so $b>1$. So solutions are given by

$$a = \frac{b}{(\sqrt{b}-1)^2}$$

for $b>1$.

g) Squaring both sides of the equation gives $$a+b+2\sqrt{ab}=a+b+\sqrt{4ab}$$ which is clearly true for all non-negative $a$ and $b$.

h) Take the RHS to the LHS of the equation and factorise out $\sqrt{a}+b$, so that

$$(\sqrt{a}+b)(\frac{c-d^2}{\sqrt{c}+d})=0$$

Then have the solution sets:

$a=b=0$ for any $c$ and $d$ non-negative not both equal to zero

or

any non-negative $a,b$ and all positive $c=d^2$.

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.