How do you solve by completing the square #a^2 -6a-5=0#?

2 Answers
Jun 8, 2018

Answer:

Solution: #a = 3+sqrt 14 and a= 3- sqrt 14#

Explanation:

#a^2-6 a-5=0 or a^2-6 a =5# or

# a^2-6 a +9 =5 +9# or

# (a-3)^2=14 or a-3 = +-sqrt 14# or

#a = 3+- sqrt 14#

Solution: #a = 3+sqrt 14 and a= 3- sqrt 14# [Ans]

Jun 8, 2018

Answer:

#a = 3+sqrt(14), a=3-sqrt(14)#

Explanation:

When we want to make a square out of a quadratic polynomial, we need to look at the b-coefficient of the polynomial (assuming #0 = ax^2+bx+c#).

We know that #(a-b)^2 = a^2 - 2ab -b^2#.
Therefore our "a" is just going to be a, and b is going to be some number for which :

# -2ab = -6a #
#-2b = -6#
#b = 3#

So our square is going to be #(a-3)^2#. However, the square equals to #a^2 - 6a + 9# and we lack the +9 to complete the square.
Luckily, we are free to add a zero to the equation anytime we want. Like this :

#a^2 - 6a + (9 - 9) - 5 = 0#
#a^2 -6a + 9 - 9 -5 = 0#
#(a-3)^2 -14 = 0#
#(a-3)^2 = 14#

Now we can square the equation, watch out for the plus/minus signs :

#|a-3| = sqrt(14)#
#a-3 = +-sqrt(14)#
#a = 3 +-sqrt(14)#

And we are finished.