Question

Solve x2-5x-4=0 by completing the square. What are the solutions?

{5+,5-}
{4+,4-}
5+sqr root of 41 /2 5-sqr root 41 /2

Answer 1

in form
ax²+bx+c=0
if a=1 then
take 1/2 of b and square it
then add that to both sides
then complete square
the move constnat to left
then sqrt both sides
then minus whatever


x²-5x-4=0
a=1 so
-5/2=-2.5, (-2.5)²=6.25
add that to both sides
x²-5x+6.25-4=6.25
factor perfect square
(x-2.5)²-4=6.25
add 4 to both sides
(x-2.5)²=10.25
sqrt both sides
x-2.5=+/-√10.25
add 2.5 to both sides
x=2.5+/-√10.25
to get into the answer form
times left side by 2/2

`x= (5+/-2 √(10.25) )/(2)`

`x= (5+/- √(4*10.25) )/(2)`

`x= (5+/- √(41) )/(2)`

answer is last option