Question

sendo p e q dois numeros positivos, tal que p-2q=4 e p+q^2= 7, qual o valor de p e q?

Answer 1

Olá,

$\begin{cases} p-2q=4~~(i)\\ p+q^2=7~~(ii)\end{cases}$

substitua ii em i:

$p+q^2=7~~\Rightarrow ~~q^2=7-p~~\Rightarrow~~q= \sqrt{7-p}~~(ii)\\\\ p-2q=4~~(i)\\ p-2\cdot( \sqrt{7-p})=4\\ -2\cdot( \sqrt{7-p})=4+p\\\\ \sqrt{7-p} = \dfrac{4+p}{-2}\\\\ \sqrt{7-p}= -\dfrac{4}{2}- \dfrac{p}{2}\\\\ ( \sqrt{7-p})^2=\left(-2- \dfrac{p}{2}\right)^2\\\\ 7-p=4+2p+ \dfrac{p^2}{4}\\\\ \dfrac{28-4p}{\not4}= \dfrac{16+8p}{\not4}+ \dfrac{p^2}{\not4}\\\\ 28-4p=16+8p+p^2\\ p^2+4p-12=0\\\\ \Delta=4^2-4\cdot1\cdot(-12)\\ \Delta=16+48\\ \Delta=64$

$p= \dfrac{-4\pm \sqrt{64} }{2\cdot1}= \dfrac{-4\pm8}{2}\begin{cases}p'=2\\ p''=-6~~(nao~serve)\end{cases}$

Vamos ao valor de q..

$q= \sqrt{7-q}\\ q= \sqrt{7-2}\\ q= \sqrt{5}$

Portanto o valor de p é 6 e q é √5
Tenha ótimos estudos ;D