Powerful quadratics

Excerpt: This problem offers a thought-provoking challenge, but one which is accessible if students are given some time to think about the problem. It is great for practising both quadratics and laws of indices.

Question

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Find all real solutions of the equation ${\large (x^2 - 7x + 11)^{(x^2 - 11x + 30)} = 1}.$
Find all real solutions of the equation ${\large (2-x^2)^{(x^2 - 3\sqrt{2}x + 4)} = 1}.$

Problem

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Find all real solutions of the equation ${\large (x^2 - 7x + 11)^{(x^2 - 11x + 30)} = 1}.$
Find all real solutions of the equation ${\large (2-x^2)^{(x^2 - 3\sqrt{2}x + 4)} = 1}.$

Suggestion

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…

Solution

There are several possibilities here.

Case 1. For any number $a$ we have $a^0=1$ , so if we solve $x^2-11x+30=0$ then this will give us some solutions.

Case 2. For any number $b$ we have $1^b=1$ , so if we solve $x^2-7x+11=1$ we will find more solutions.

Case 3. We also have that $(-1)^{2b}=1$ for any number $b$ . So we can solve $x^2-7x+11=-1$ , and then see whether the numbers we get lead to $x^2-11x+30$ being an even number.

These are the only options, so by dealing with each case in turn we’ll find all the solutions.

For Case 1, we get $x^2-11x+30=(x-6)(x-5)=0,$ so $x=5$ and $x=6$ are solutions.

For Case 2, we solve $x^2-7x+10=(x-2)(x-5)=0,$ so we have an additional solution, $x=2$ .

Finally, for Case 3 we have that

x^2-7x+12=(x-3)(x-4)=0,

so we must check:

\begin{align*} 3^2-11\cdot 3+30=6 \quad \checkmark\\ 4^2-11\cdot 4 +30=2 \quad \checkmark \end{align*}

are both even, so $x=3$ and $x=4$ are also solutions.

So the real numbers satisfying the equation are $x=2,3,4,5, 6.$

There are again three cases to consider:

Case 1. $x^2-3\sqrt{2}x+4=0$ ,

Case 2. $2-x^2=1$ ,

Case 3. $2-x^2=-1$ and $x^2-3\sqrt{2}x+4$ is even.

In Case 1, $x^2-3\sqrt{2}x+4=(x-\sqrt{2})(x-2\sqrt{2})=0,$ and so we have that $x=\sqrt{2}$ and $x=2\sqrt{2}$ are solutions. However, when $x=\sqrt{2}$ , we end up with $0^0$ .

The quantity $0^0$ is a bit problematic. For some mathematicians there are good reasons to define $0^0$ to be $1$ , whereas other mathematicians prefer to say that $0^0$ is undefined.

So we may choose to discard $x=\sqrt{2}$ as a solution.

In Case 2, we have that $x^2-1=(x+1)(x-1)=0,$ so $x=1$ and $x=-1$ are also solutions.

Solving Case 3, we get $x^2-3=(x+\sqrt{3})(x-\sqrt{3})=0,$ and so $x=\pm\sqrt{3}$ are potential solutions. However, we have that $\sqrt{3}^2\pm3\sqrt{2}\sqrt{3}+4=7\pm3\sqrt{6}$ is not divisible by two (it isn’t even an integer!) so $x=\pm\sqrt{3}$ are not solutions.

So the values of $x$ satisfying the equation are $x=-1, 1, 2\sqrt{2}.$