Respuesta :
[tex] x^2 - 8x + 5 = 0\\
x^2-8x+16-11=0\\
(x-4)^2=11\\
x-4=\sqrt{11} \vee x-4=-\sqrt{11}\\
x=4+\sqrt{11} \vee x=4-\sqrt{11} [/tex]
First, let's move the 5 to the other side by subtracting both sides by 5. That leaves us with:
x^2 - 8x = -5
The constant on the left side to make a binomial square should be (8/2)^2 = 16. Add 16 to both sides and you get: x^2 - 8x + 16 = 11
(x - 4)^2 = 11
Square root both sides:
(x - 4) = +/- sqrt (11)
Add 4 to both sides:
x = +/- sqrt (11) + 4
The 4 should be outside the square root.