Question

Somente uma das equações abaixo tem as raízes 2 e 3. Qual é? *
a) 2x² - 5x + 6 = 0
b) x² - 5x – 6 = 0
c) x² + 5x + 6 = 0
d) – x² + 5x – 6 = 0

Answer 1

$a)~x= \frac{-b\pm \sqrt{b^2-4.a.c}}{2.a} \\ \\ x= \frac{5\pm \sqrt{(-5)^2-4.2.6}}{2.2} \\ \\ x= \frac{5\pm \sqrt{25-24}}{4} \\ \\ x= \frac{5\pm \sqrt{1}}{4} \\ \\ x= \frac{5\pm1}{4}$

$x'= \frac{5+1}{4}= \frac{6}{4}:2= \frac{3}{2}$

$x''= \frac{5-1}{4}= \frac{4}{4}=1$

$b)~x= \frac{-b\pm \sqrt{b^2-4.a.c}}{2.a} \\ \\ x= \frac{5\pm \sqrt{(-5)^2-4.1.(-6)}}{2.1} \\ \\ x= \frac{5\pm \sqrt{25+24}}{2} \\ \\ x= \frac{5\pm \sqrt{49}}{2} \\ \\ x= \frac{5\pm7}{2}$

$x'= \frac{5+7}{2}= \frac{12}{2}=6 \\ \\ x''= \frac{5-7}{2}= \frac{-2}{2}=-1$

$c)~x= \frac{-b\pm \sqrt{b^2-4.a.c}}{2.a} \\ \\ x= \frac{-5\pm \sqrt{5^2-4.1.6}}{2.1} \\ \\ x= \frac{-5\pm \sqrt{25-24}}{2} \\ \\ x= \frac{-5\pm \sqrt{1}}{2} \\ \\ x= \frac{-5\pm1}{2}$

$x'= \frac{-5+1}{2}= \frac{-4}{2}=-2 \\ \\ x''= \frac{-5-1}{2}= \frac{-6}{2}=-3$

$d)~x= \frac{-b\pm \sqrt{b^2-4.a.c}}{2.a} \\ \\ x= \frac{-5\pm \sqrt{5^2-4.(-1).(-6)}}{2.(-1)} \\ \\ x= \frac{-5\pm \sqrt{25-24}}{-2} \\ \\ x= \frac{-5\pm \sqrt{1}}{-2} \\ \\ x= \frac{-5\pm1}{-2}$

$x'= \frac{-5+1}{2}= \frac{-4}{-2}=2 \\ \\ x''= \frac{-5-1}{-2}= \frac{-6}{-2}=3$


Alternativa D.