Question

solve the equation by completing the square.
x^2-7x-4=0
a. 7.53, -0.53
b. -7.53, -0.53
c. -7.53, 0.53
d. 7.53, 0.53

Answer 1

Well since the coefficient of the x^2 term is a 1, we can start by taking the coefficient of the x term and divide it by 2 so that we have (7/2), now we square that getting 49/4. we are going to add that value to the left hand side giving us X^2 -7x + 49/4 -4 =0.
But we have unbalance the equation so we must add that value to the right hand side. Giving up x^2 - 7x + 49/4 -4 = 49/4. we now add 4 to both sides giving use
x^2 - 7x + 49/4 = 49/4 + 4. The expression on the left of the equal sign is now a perfect square and can be written as (x - 7/2)^2 = 49/4 - 4.
Do the subtraction on the right side getting (x - 7/2)^2 = 33/4
Solve by taking the square root of both sides getting:
x - 7/2 =Sq rt 33/2 \[x =\pm \sqrt{33}/2 + 7/2 \]