Powerful Quadratics

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:

• $ax^2+bx+c=1$, 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 $n$.

Now to solve the problems:

$i)$ 

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 $2 \sqrt{2}$.