Question

$x {}^{2} - 2(a + 1)x + 4a = 0$
resolver com a e R

Answer 1

$x^2 - 2(a+1)x+4a=0\\\\ \Delta = b^2- 4ac\\ \Delta = [-2(a+1)]^2-4.(1).(4a)\\ \Delta = (-2a-2)^2-16a\\ \Delta = 4a^2+8a+2-16a\\ \Delta = 4a^2-8a+1\\ \Delta = (2a-1)^2\\\\ x=\dfrac{-b\pm \sqrt{\Delta} }{2a}\\\\ x=\dfrac{-[-2(a+1)]\pm \sqrt{(2a-1)^2} }{2}\\\\ x=\dfrac{[2(a+1)]\pm(2a-1) }{2}\\\\ x=\dfrac{(2a+2)\pm(2a-1) }{2}\\\\\\ x'=\dfrac{(2a+2)+(2a-1) }{2} = \dfrac {4a+1}{2}\\\\\\ x'=\dfrac{(2a+2)-(2a-1) }{2} = \dfrac {3}{2}$

Como temos umas das raízes sem incógnitas, podemos descobrir o valor de a, substituindo na equação original:

$x^2 - 2(a+1)x+4a=0\\\\ (\dfrac{3}{2})^2 - 2(a+1)(\dfrac{3}{2}) + 4a = 0\\\\ \dfrac{9}{4} - \dfrac{6a}{2} -\dfrac{6}{2}+ 4a = 0\\\\ mmc(4,2,1)=4\\\\ \dfrac{9-12a-12+16a=0}{4}\\\\ 9-12a-12+16a=0\\\\ 4a -3 = 0\\\\ 4a = 3\\\\ a=\dfrac{3}{4}$

Como descobrimos o valor de a, podemos calcular o valor da primeira raiz:

$x' = \dfrac {4a+1}{2}\\\\ x' = \dfrac {4(\dfrac{3}{4})+1}{2}\\\\ x' = \dfrac {3+1}{2}\\\\ x' = \dfrac {4}{2}\\\\ x' = 2$

Qualquer dúvida só chamar!

=)