Matematica - Ecuatii si inecuatii. Functii

Ecuatii si inecuatii. Functii - Intrebarea 1

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Să se determine soluțiile reale ale ecuației: $square root of x squared minus x minus 2 end root equals 2$ .

x element of open curly brackets negative 1 semicolon 2 close curly brackets

x equals negative 2

x element of open curly brackets negative 2 semicolon 3 close curly brackets

Raspunsul este corect.

Raspunsul este gresit.

Explicatie:

Răspuns corect: C

Punem condițiile de existență: $x squared minus x minus 2 greater or equal than 0 left right double arrow x element of left parenthesis negative infinity comma negative 1 right square bracket union left square bracket 2 comma plus infinity right parenthesis$ .

$square root of x squared minus x minus 2 end root equals 2 vertical line squared left right double arrow x squared minus x minus 6 equals 0 left right double arrow x element of open curly brackets negative 2 semicolon 3 close curly brackets.$

Ambele soluții convin, deoarece apar'in domeniului $left parenthesis negative infinity comma negative 1 right square bracket union left square bracket 2 comma plus infinity right parenthesis$ .