Question

Sendo U=R , resolva as equações irracionais

Answer 1

1ª

$\text x = \sqrt{6-\text x} \\\\ \underline{\text{elevando ao quadrado dos dois lados}}: \\\\ \text x^2=\sqrt{(6-\text x)^2} \\\\ \text x^2 = | 6-\text x| \\\\ \text{temos : } \\\\ \text x^2 = 6 -\text x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ou} \ \ \ \ \ \ \ \ \text x^2=-6+\text x \\\\\text x^2+\text x-6=0 \ \ \ \ \ \ \ \ \ \ \text{ou} \ \ \ \ \ \ \ \ \ \ \text x^2-\text x+6=0$

$\displaystyle \text x = \frac{-1\pm\sqrt{1^2-4.1.(-6)}}{2.1} \ \ \ \ \ \ \ \ \ \ \text {ou} \ \ \ \ \ \ \ \ \ \text x = \frac{-(-1)\pm\sqrt{(-1)^2-4.6}}{2} \\ \text x = \frac{-1\pm\sqrt{25}}{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{raiz negativa nao convem}) \\\\ \text x =\frac{-1\pm5}{2} \\\\ \text x = \frac{-1+5}{2} \to \boxed{\text x = 2} \\ \text x = \frac{-1-5}{2}\to \boxed{\text x = -3} \text{(FALSO)}$

Solução :

$\huge\boxed{ \text x = 2}\checkmark$

2ª

$\sqrt{\text x+3} = \text x-3 \\\\ \underline{\text{elevando ao quadrado dos dois lados}}: \\\\ \sqrt{(\text x+3)^2}=(\text x-3)^2 \\\\ |\text x+3| = \text x^2-6\text x+9 \\\\ \text{casos :} \\\\ \text x^2-6\text x+9=\text x+3 \ \ \ \ \ \ \ \ \ \text{ou} \ \ \ \ \ \ \ \ \ \text x^2-6\text x+9=-\text x-3 \\\\\text x^2-7\text x+6=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ou} \ \ \ \ \ \ \ \ \ \text x^2-5\text x+12=0$

$\displaystyle \text x=\frac{-(-7)\pm\sqrt{(-7)^2-4.6}}{2.1} \ \ \ \ \ \ \text{ou} \ \ \ \ \ \ \ \text x=\frac{-(-5)\pm\sqrt{(-5)^2-4.12}}{2.1} \\\\\text x = \frac{7\pm\sqrt{49-24}}{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ou} \ \ \ \ \ \ \ \ \ \text{( raiz negativa nao convem )} \\\\ \text x = \frac{7\pm5}{2} \\\\ \text x = \frac{7+5}{2} \to \boxed{\text x = 6}\ \text{(VERDAEIRO)} \\ \text x = \frac{7-5}{2} \to \boxed{\text x = 1} \ \text{(FALSO)}$

Solução :

$\huge\boxed{\text x = 6}\checkmark$