How do you factor #x^ { 2} - 17x + 65= 0#?

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Answer 1 · 2017-02-08T12:09:01

#(x-17/2-sqrt29/2)(x-17/2+sqrt29/2)=0#

Solved, # x = (17 \pm sqrt(29)) / 2 #

We start with:

#x^2-17x+65=0#

We're now going to look for two numbers that multiply to 65 and add to #-17#. To do this, let's consider factors of 65:

#-1, -5, -13,-65#

and there aren't any that get us there.

Let's factor using the Quadratic Formula:

# x = (-b \pm sqrt(b^2-4ac)) / (2a) #

# x = (17 \pm sqrt((-17)^2-4(1)(65))) / (2(1)) #

# x = (17 \pm sqrt(289-260)) / 2 #

# x = (17 \pm sqrt(29)) / 2 #

# x = 17/2+sqrt(29) / 2 , 17/2-sqrt29/2#

And so we can factor:

#(x-17/2-sqrt29/2)(x-17/2+sqrt29/2)=0#