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# Powerful Quadratics

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### Discriminating

### Factorisable Quadratics

### Which Quadratic?

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Age 16 to 18

- Problem
- Getting Started
- Student Solutions

Thank you to Maria from Hills Road Sixth Form College, Eleanor from The King's School in Macclesfield, Abi, and Rhombusta for submitting solutions to this problem!

Here's Abi's solution:

For equations of the form $(ax^2+bx+c)^{(dx^2+ex+f)}=1$, where $a,b,c,d,e,f$ and $x$ are real, there are three possibilities:

$i)$

Case $1$: $x^2-7x+11=1$ factorises as $(x-5)(x-2)=0$ so $x=5$ or $x=2$.

Case $2$: $x^2-11x+30=0$ factorises as $(x-5)(x-6)=0$, so $x=6$ or $x=5$.

Case $3$: $x^2-7x+11=-1$ factorises as $(x-3)(x-4)=0$. As $3^2-11(3)+30=6$ and $4^2-11(4)+30=2$ are both even, $x=3$ or $x=4$ are solutions.

So the solutions are $x=2,3,4,5$ and $6$.

$ii)$

Case $1$: $2-x^2=1$ factorises as $(1-x)(1+x)=0$ so $x=1$ or $x=-1$.

Case $2$: $x^2-3 \sqrt{2}x+4=0$ factorises as $(x-2 \sqrt{2})(x- \sqrt{2})$, so $x=2 \sqrt{2}$ or $x = \sqrt{2}$.

Case $3$: $2-x^2=-1$ factorises as $(\sqrt{3}-x)(\sqrt{3}-x)$, but as $\sqrt{3}^2-3 \sqrt{2} \sqrt{3} +4$ and $ \sqrt{3}^2-3 \sqrt{2} \sqrt{3} + 4$ are not even integers, these are invalid.

So the solutions are $x=-1,1, \sqrt{2}$ and $2 \sqrt{2}$.

Here's Abi's solution:

For equations of the form $(ax^2+bx+c)^{(dx^2+ex+f)}=1$, where $a,b,c,d,e,f$ and $x$ are real, there are three possibilities:

- $ax^2+bx+c=1$, as $1^n=1$ for any $n$
- $dx^2+ex+f=0$, as $n^0=1$ for any $n$
- $ax^2+bx+c=-1$, and $dx^2+ex+f$ is an even integer, as $(-1)^{(2n)}=1$ for any integer $n$.

$i)$

Case $1$: $x^2-7x+11=1$ factorises as $(x-5)(x-2)=0$ so $x=5$ or $x=2$.

Case $2$: $x^2-11x+30=0$ factorises as $(x-5)(x-6)=0$, so $x=6$ or $x=5$.

Case $3$: $x^2-7x+11=-1$ factorises as $(x-3)(x-4)=0$. As $3^2-11(3)+30=6$ and $4^2-11(4)+30=2$ are both even, $x=3$ or $x=4$ are solutions.

So the solutions are $x=2,3,4,5$ and $6$.

$ii)$

Case $1$: $2-x^2=1$ factorises as $(1-x)(1+x)=0$ so $x=1$ or $x=-1$.

Case $2$: $x^2-3 \sqrt{2}x+4=0$ factorises as $(x-2 \sqrt{2})(x- \sqrt{2})$, so $x=2 \sqrt{2}$ or $x = \sqrt{2}$.

Case $3$: $2-x^2=-1$ factorises as $(\sqrt{3}-x)(\sqrt{3}-x)$, but as $\sqrt{3}^2-3 \sqrt{2} \sqrt{3} +4$ and $ \sqrt{3}^2-3 \sqrt{2} \sqrt{3} + 4$ are not even integers, these are invalid.

So the solutions are $x=-1,1, \sqrt{2}$ and $2 \sqrt{2}$.

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.

In this activity you will need to work in a group to connect different representations of quadratics.