Question

(Pucrj) Quanto vale a soma de todas as soluções reais da equação abaixo?
(5^x)^2-26*5^x+25=0

Answer 1

Olá,

aplique as propriedades da exponenciação:

$\mathsf{(5^x)^2-26\cdot5^x+25=0}\\\\ \mathsf{5^x=y}\\\\ \mathsf{y^2-26y+25=0}\\\\ \mathsf{\Delta=(-26)^2-4\cdot1\cdot25}\\ \mathsf{\Delta=676-100}\\ \mathsf{\Delta=576}\\\\ \mathsf{y= \dfrac{-(-26)\pm \sqrt{576} }{2\cdot1}= \dfrac{26\pm24}{2} }\begin{cases}\mathsf{y_1= \dfrac{26-24}{2}= \dfrac{2}{2} =1 }\\\\\mathsf{y_2= \dfrac{26+24}{2}= \dfrac{50}{2} =25 }\end{cases}\\\\\\ \mathsf{5^x=y}$

$~\mathsf{5^x=1~~~~~~~~~~~~~5^x=25}\\ \mathsf{\not5^x=\not5^0~~~~~~~~~~~\not5^x=\not5^2}\\ ~~~\mathsf{x_1=0}~~~~~~~~~~~~~~\mathsf{x_2=2}$

Portanto a soma de todas as soluções é:

$\mathsf{x_1+x_2=0+2}\\\\ \huge\boxed{\mathsf{x_1+x_2=2}}$

Answer 2

5^x=y

y^2-26y+25=0

∆=b^2-4.a.c

∆=(-26)^2-4.(25)

∆=676-100

∆=576

y1=26+24/2

y1=50/2

y1=25

y2=26-24/2

y2=2/2

y2=1


y1=25 e y2=1

5^x=y

5^x=5^2

x=1

5^x=1

5^x=5^0

x=0

X1+X2=2+0

X1+X2=2

espero ter ajudado!

boa noite!