How do you solve #w ^ { 2} - 6w + 9= 13#?

2 Answers
Nov 3, 2016

Answer:

#w = 3+-sqrt(13)#

Explanation:

Given:

#w^2-6w+9 = 13#

Note that the left hand side is already a perfect square trinomial, so we have:

#(w-3)^2 = 13#

Taking the square root of both sides, allowing for both possible square roots, we have:

#w-3 = +-sqrt(13)#

Then add #3# to both sides to get:

#w = 3+-sqrt(13)#

Nov 3, 2016

Answer:

#w=3+-sqrt13#

Explanation:

#w^2-6w+9=13#
Rearrange
#w^2-6w-4=0#
Either use the formula #w=(-b+-sqrt(b^2-4ac))/(2a)#
Where #aw^2+bw+c=0#
#w=(6+-sqrt(36-4*1*-4))/2#
#w=(6+sqrt(36+16))/2# or #w=(6-sqrt(36+16))/2#
#w=(6+sqrt52)/2=3+sqrt13#
Or
#w=3-sqrt13#

The other way is completing the square ( which is how the formula is derived)
#w^2-6w-4=0#
#(w-3)^2=w^2-6w+9#
So #w^2-6w-4# can be written #(w-3)^2-13=0#
#(w-3)^2=13#
Take the square root of both side
#w-3=+-13#
#w=3+-13#