How do you factor #x^ { 2} + 13x - 169#?

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Answer 1 · 2018-03-12T17:39:55

There are no whole number factors.

#(x-8.11)(x+8.11)# to 2 decimal places

using the formula

Set #x^2+13x-169=0# then we have:

#y=ax^2+bx+c color(white)("ddd") ->color(white)("ddd")x=(-b+-sqrt(b^2-4ac))/(2a)#

Where #a=1, b=13, c=-169#

#x=(-13+-sqrt(13^2-4(1)(-169)))/(2(1))#

#x=-13/2+-sqrt(854)/2#

Prime factors of 854 are 2,7 and 61

So the exact answer is #x=-13/2+-sqrt(854)/2#

The approximate answer is: #x=-8.11 and x=+8.11# 2 dp

color(white)("d")