PMO 4

Algebra Level 2

Solve the equation:

$\large (2-x^2)^{x^2-3\sqrt{2}x+4} = 1$

+2, -2, 2\sqrt{2}

+4, -4, 2\sqrt{2}

+3, -3, 2\sqrt{2}

+1, -1, 2\sqrt{2}

No solution exists

1 solution

Sir isaac Newton
Nov 10, 2017

an exponential is equal to 1 either if the base is equal to 1 or if the exponent is equal to 0 and the base is not. so for the base equal to 1 we have $(2-x^2)=1$ , so $x=+1, x=-1$ . for the exponent equal to 0 we have $x^2-3\sqrt{2}x+4=0$ the two roots are $\sqrt{2}$ and $2\sqrt{2}$ but since $\sqrt{2}$ makes the base 0 we can't accept it, thus the solutions will be $1$ , $-1$ and $2\sqrt{2}$ .

Thank you Sir Issac for sharing your solution

Hana Wehbi - 3 years, 7 months ago

Read this about 0^0: https://brilliant.org/wiki/what-is-00/ It's hard to say whether we may exclude the case of $\sqrt{2}$ .

Peter van der Linden - 3 years, 7 months ago

I will but the problem is not original, so l am the author considered all possibilities (maybe).

Hana Wehbi - 3 years, 7 months ago

PMO 4

1 solution

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