# Powerful Quadratics

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.

(i) Find all real solutions of the equation

$$(x^2âˆ’7x+11)^{(x^2âˆ’11x+30)}=1.$$

(ii) Find all real solutions of the equation

$$(2âˆ’x^2)^{(x^2âˆ’3\sqrt{2}x+4)}=1.$$

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Find all real solutions of the equation

$$(x^2âˆ’7x+11)^{(x2âˆ’11x+30)}=1.$$

If you were told that two numbers satisfied $a^b=1$, could you write down the possible numbers that a and b could be?

Can you apply the same logic to this question? You might have to think hard to find all the solutions to the given equation. There are more than three solutions”¦

$$(x^2âˆ’7x+11)^{(x2âˆ’11x+30)}=1.$$

If you were told that two numbers satisfied $a^b=1$, could you write down the possible numbers that a and b could be?

Can you apply the same logic to this question? You might have to think hard to find all the solutions to the given equation. There are more than three solutions”¦

**This is an Underground Mathematics resource.**

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 $2 \sqrt{2}$.