Liczby -2√2 i 5√2 są miejscami zerowymi funkcji, zatem można wyznaczyć postać iloczynową:
[tex]f(x) = a(x + 2 \sqrt{2})(x - 5 \sqrt{2} ) \\ [/tex]
[tex] \frac{f( \sqrt{2}) }{f(2 \sqrt{2}) } = \frac{a( \sqrt{2} + 2 \sqrt{2})( \sqrt{2} - 5 \sqrt{2} )}{a(2 \sqrt{2} + 2 \sqrt{2})(2 \sqrt{2} - 5 \sqrt{2} )} \\ \\ \frac{f( \sqrt{2}) }{f(2 \sqrt{2}) } = \frac{(3\sqrt{2})(- 4\sqrt{2} )}{(4\sqrt{2})( - 3\sqrt{2} )} \\ \\ \frac{f( \sqrt{2}) }{f(2 \sqrt{2}) } = 1 \\ [/tex]
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Liczby -2√2 i 5√2 są miejscami zerowymi funkcji, zatem można wyznaczyć postać iloczynową:
[tex]f(x) = a(x + 2 \sqrt{2})(x - 5 \sqrt{2} ) \\ [/tex]
[tex] \frac{f( \sqrt{2}) }{f(2 \sqrt{2}) } = \frac{a( \sqrt{2} + 2 \sqrt{2})( \sqrt{2} - 5 \sqrt{2} )}{a(2 \sqrt{2} + 2 \sqrt{2})(2 \sqrt{2} - 5 \sqrt{2} )} \\ \\ \frac{f( \sqrt{2}) }{f(2 \sqrt{2}) } = \frac{(3\sqrt{2})(- 4\sqrt{2} )}{(4\sqrt{2})( - 3\sqrt{2} )} \\ \\ \frac{f( \sqrt{2}) }{f(2 \sqrt{2}) } = 1 \\ [/tex]