SOLUTION: solve by completing the square: 2x^2-x-5=0

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: solve by completing the square: 2x^2-x-5=0      Log On


   

Question 137172: solve by completing the square: 2x^2-x-5=0
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
2x^2 - x - 5 = 0
:
When you complete the square, you want the coefficient of x^2 to be 1,
divide the equation by 2, to accomplish this
x^2 - 1%2F2x - 5%2F2 = 0
:
Add 5%2F2 to both sides, leave a place for the value we need to complete the square
x^2 - 1%2F2x + ___ = 5%2F2
:
Find the third term by dividing the coefficient of x by 2 and squaring it:
Here that would be; %281%2F4%29%5E2 which is 1%2F8, we now have:
x^2 - 1%2F2x + 1%2F8 = 5%2F2 + 1%2F8; we have to add 1%2F8to both sides
x^2 - 1%2F2x + 1%2F8 = 20%2F8 + 1%2F8; find a common denominator so we cam add the fractions
x^2 - 1%2F2x + 1%2F8 = %2821%29%2F8
We have perfect square which is:
(x - 1%2F4)^2 = %2821%29%2F8
Find the square root of both sides:
x - 1%2F4 = +/-sqrt%2821%2F8%29
we can extract the sqrt of 4 in the denominator;
x - 1%2F4 = +/-%281%2F2%29%2Asqrt%2821%2F2%29
add 1/4 to both sides
x = 1%2F4+/-%281%2F2%29%2Asqrt%2821%2F2%29
or we can write it
x = 1%2F4+/-%282%2F4%29%2Asqrt%2821%2F2%29
Put it all over a denominator of 4
x = %281+%2B+2%2Asqrt%2821%2F2%29%29%2F4
and
x = %281+-+2%2Asqrt%2821%2F2%29%29%2F4
Didthishelp?